PB
i
Arise with
Mathematics
Students’ Book 3
Johns J. Mwale
Numeri Chimalizeni
ii
iii
CLAIM Limited
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P.O. Box 503
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Malawi
© CLAIM 2014
All rights reserved. No part of this publication may be reproduced, stored
in a retrieval system or transmitted in any form, electronic, photocopying,
recording, mechanical or otherwise except with prior written permission from
the publisher.
Editor : Blanzio Dayton
Proofreader : Thokozani Eston Mkwate
Designed by : Emmanuel Chikaonda
Joe Kima Phulusa
First Published 2014
ISBN : 978-99960-35-69-2
ii
iii
Contents
Unit 1 Quadratic equation . . . . . . . . . . . . . . 1
Unit 2 Irrational numbers . . . . . . . . . . . . . . 23
Unit 3 Circle geometry (chord properties) . . . . . . 43
Unit 4 Algebraic fractions with linear or quadratic
denominators . . . . . . . . . . . . . . 59
Unit 5 Sets . . . . . . . . . . . . . . . . . . . . 77
Unit 6 Mapping and functions . . . . . . . . . . 105
Unit 7 Circle geometry (angle properties) . . . . . . . 125
Unit 8 Transformation . . . . . . . . . . . . . . 145
Unit 9 Change of subject of aformula. . . . . . . . . . 185
Unit 10 Exponential and Logarithmic Equations . . . . 194
Unit 11 Trigonometry . . . . . . . . . . . . . . . . 215
Unit 12 Similarity . . . . . . . . . . . . . . . . . . 251
Unit 13 Coordinate Geometry . . . . . . . . . . . . 269
Unit 14 Variations . . . . . . . . . . . . . . . . . . 291
Unit 15 Graphs of quadratic Functions. . . 316
Unit 16 Inequalities . . . . . . . . . . . . . . . . 341
Unit 17 Statistics . . . . . . . . . . . . . . . . . . 357
iv
1
Dedication
This book is dedicated to Benjamin Chimalizeni and Clever Mwale who are the
last born to Mr Chiamilzeni and Mr Mwale respectively. May you be inspired
by this work as you journey in this life. God be with you and your brothers and
sisters and not forgetting your mothers.
Acknowledgements
We would like to thank God for protecting and keeping us of good health and
seeing us through the entire period of writing this book.
Thanks also go to Mrs Janet Mwale and Mrs Gladys Chimalizeni for their
support and encouragement during the time the book was being written. We
know, our time was dedicated to this work rather than you but you were patient
with us. May God bless you all.
We would also like to thank Mr Chisamba, Mr Yohane, Mr Muphuwa , Mr Bob
Tsapa and other members of CLAIM for their encouragement and support.
We were about to give up but you came forward and encouraged us. May the
Almighty God bless you.
We also extend our thanks to Mr Emmanuel Chikaonda, Mr Thokozani Mkwate
for their wonderful work and making the book look like this.
Our gratitudes also go to the Head teachers and staff of Robert Blake, Madisi,
and Mwansambo secondary schools for their moral support and encouragement.
We are proud of you.
May we also thank all those we consulted and assisted in the development of
this book and not forgetting Pastor Benjamin Makoto of Dowa for his guidance.
May God bless you. We hope all those who use the book will nd it useful.
iv
1
Unit
1
QUADRATIC EQUATIONS
In your JCE Mathematics you
learnt about quadratic expressions.
You learnt how to factorise and
expand such expressions. Recall that
expanding and factorising of quadratic
expressions are opposite operations.
In this unit you will learn about
quadratic equations and you will apply
the skills you acquired in your JCE
Mathematics. Specically you will
solve quadratic equations by making
use of a number of methods such as
factorisation, completing squares, and
using a quadratic formula. You will
nally apply quadratic equations to
solving everyday problems.
Denition: Quadratic
equation
Factorising quadratic
expressions
A quadratic expression is any
expression in mathematics which
takes the form ax
2
+ bx + c where
a, b and c are real numbers and
a ≠ 0 .You will learn more about
expressions of this form in this unit.
Activity 1:
Identifying quadratic
expressions
Which of the following expressions are
quadratic expressions?
a) x
2
+ 2x + 1
b) x
3
+ 3x + 2
c) 2x
2
+ 1
d) x
2
– 1
e) x + xy + 2
f) x
4
+ 3x + 2
2
3
Understanding factorisation
Denition: Factorisation
In Mathematics, factorisation is the decomposing of a number or
an expression into a product of other numbers or factors, which
when multiplied together will you give the original number or
expression.
For example, the number 20 can be factorised as 4 x 5 or 2 x10.
And the expression x
2
- 4 can be factorised as
(x + 2) (x – 2). In both of these examples, a product of simpler
numbers or expressions is obtained.
Have you ever asked yourself why you need to learn
factorisation? Or have you wondered why you should bother
yourself factorising an expression at all?
The aim of factoring or factorisation is usually to reduce an
expression to some simple expressions.
Before you learn how to factorise, it is important for you to know
different types of quadratic expressions. Remember that in a
quadratic expression, the highest power of the letters is a 2.
Types of quadratic expressions
Quadratic expressions are classied basing on how many terms
are in the expression:
a) Quadratic Monomial Expression
A monomial is an expression in algebra which contains
one term. A quadratic monomial expression has one term
only. Examples of quadratic monomials include x
2
, 2x
2
, ½ x
2
,
5x
2
, and many others. Think about ve more examples of
quadratic monomials.
a) Quadratic Binomial Expressions
A binomial is an expression in algebra which contains two
terms. `bi` means `2`. A quadratic monomial expression has
two terms in it. Examples of quadratic monomials include
2
3
x
2
+ 2x, 2x
2
+3, 5x
2
+ 7 and many others. Think about ve
more examples of quadratic binomial expressions.
b) Quadratic Trinomial Expressions
A trinomial is an algebraic expression with three terms.
`Tri` means `3`. A quadratic trinomial has three terms in
it. The following expressions are examples of trinomials:
i) d
2
+7d + 10
ii) x
2
+ 11x + 10
iii) c
2
+ 8c + 15
iv) x
2
+ 5x + 6
v) f
2
+ 7f + 6
vi) g
2
+ 10g – 24
vii) x
2
+ 2x 24
Factorising quadratic expressions
Factoring trinomials of the type x² + bx + c
The standard format for a trinomial is
cbxax ++²
. In this
section, you will factorise trinomials where the coefcient of the
x
2
term is 1, (a = 1).
When we factor trinomials of the form
cbxx ++²
you are nding
2 binomials that will multiply out to give this initial polynomial.
Use the following steps to factor out this type of trinomials:
1) Draw two sets of parentheses for the 2 binomials you are
looking for: x
2
+6x+3+5= ( )( )
2) Put an x on the rst terms of the two binomials:
x
2
+2xy-15y
2
= (x-3y) (x+5y).
3) Find 2 integers whose product is c and whose sum is b. It
helps to list all the possible factors of c and check to see
which set of factors sum up to.
4) Put these integers on the last terms of the two binomials.
4
5
You can read the steps above, over and over again until you
fully understand what we are saying about factorising quadratic
trinomials.
Examples:
Factorising quadratic trinomials
Factorise the following quadratic trinomials:
a)
56² ++ xx
b)
2
920aa
c)
2
421mm+ −
d) t
2
– 15 – 2t
e)
22
215
x xy y
+ −
f) 2a
2
– 16a + 32
Answers:
a) x
2
+ 6x +5 =
( )( )
=
( )( )
xx
=
Factoring Trinomials in the form ax² + bx + c
In this section, you will factor out trinomials where the coefcient of the
2
x
term is a number other than 1(a ≠ 1).
When you factor trinomials of the form
cbxax ++²
you are also nding 2
binomials that will multiply out to give the initial polynomial. There are 2
methods that you can use to factor this type of trinomials.
1. Factor with FOIL (Trial and Error)
2. Factor by Grouping – also called the Master Product Method or the
“AC” Method
Denitions
Monomial is an algebraic expression containing one term. eg 2x
2
4
5
Binomial is an algebraic expression containing two terms, for
example 5x
2
+ 2.
Trinomial is an algebraic expression containing three terms, for
example x
2
+ 5x + 6.
Factorising quadratic expressions
Example 1:
Factorisation
Factorise the following;
(a) x
2
+ 7x + 12
Solution;
Start by writing (x ) (x )
Find numbers which multiply to give +12 and which add to
give + 7
Which numbers are these?
∴ x
2
+ 7x + 12 = (x + 3) (x + 4)
The process of nding the numbers will not be very easy for
a start but with more practice, you will nd it not too hard.
(b) x
2
+ 4x – 12
Solution;
Here the two numbers must multiply to give –12, this
means one number is positive and the other negative. The
same two numbers should add to give +4. The two numbers
are +6 and –2.
∴ x
2
+ 4x – 12 = (x + 6) (x – 2)
(c) x
2
– 11x –12
6
7
Solution:
Here the two numbers should multiply to give -12 and again
one of the numbers is positive and the other negative. Since
the two numbers add to -11, the smaller is positive and the
larger is negative. The numbers are +1 and –12
∴ x
2
–11x –12 = (x + 1) (x – 12)
(d) h
2
–8h +12
Solution
Here the two numbers should multiply to give +12 and
add to give –8, this means the two numbers must both be
negative the numbers are –2 and –6
∴h
2
–8h + 12 = (h – 2) (h – 6)
Exercise 1a
Factorise the following quadratic expressions.
1. d
2
+7d + 10
2. x
2
+11x + 10
3. c
2
+ 8c + 15
4. x
2
+ 5x + 6
5. f
2
+ 7f + 6
6. g
2
+ 10g –24
7. x
2
+ 2x – 24
8. b
2
+ 5b – 24
9. x
2
+ 23x – 24
10. m
2
–23m–-24
11. v
2
+12v + 36
12. x
2
– 2x – 24
13. y
2
–10y – 24
14. m
2
–m–30
15. g
2
–6g – 40
6
7
The expressions you have looked at so far have a co-efcient of
1 on the square term. Sometimes the co-efcient is more than
1.
Activity 2:
Coefcients
a. What is the difference between the coefcient of x
2
in these
two expressions
(i) x
2
– 37x + 36 and
(ii) 2x
2
– 17x – 10
a. Factorise 6x
2
–17x – 10
Compare your work with friend and the examples given
below.
Example 2:
Factorisation
Factorise the following:
(a) 2x
2
+ 11x + 12
Solution
To make 2x
2
we require one bracket to contain 2x and the
other x i.e.
(2x ) (x ). We should look for two numbers which
multiply to give +12 and when put in the two brackets, they
should multiply with 2x and x and their sum simply to give
11x, by inspection the rst number is 3 and the second is 4.
i.e. 2x
2
+ 11x + 12 = (2x + 3 ) (x + 4)
(b) 4x
2
+12x +9
Trying (4x ) (x ) does not work and we have to split the
4.
8
9
i.e. (2x ) (2x )
And the numbers are +3 and +3
∴ 4x
2
+ 12x + 9 = (2x + 3) (2x + 3)
= (2x + 3)
2
Exercise 1b
Factorise the following expressions
1. 2b
2
+ 3b + 1
2. 2t
2
+7t + 6
3. 3q
2
+ 8q + 4
4. 3x
2
+ 13x + 4
5. 2d
2
+ d – 6
6. 3x
2
+11x – 4
7. 9y
2
–6y + 1
8. 5x
2
– 13x – 6
9. 4w
2
+ 17w – 15
10. 6x
2
–17x – 10
11. a
2
b
2
+ 7ab + 10
12. 2u
2
v
2
+ uv − 6
13. 3t
2
+5st + 2s
2
14. m
2
n
2
+ 4mn − 21
15. 35 – 2u – u
2
16. 35 + 30d −5d
2
17. x
2
− 2xy − 15y
2
18. x
2
y
2
– xy – 30
19. 10p
2
– 43p + 45
20. x
2
+ 16xy − 36y
2
8
9
Now look at these two expressions;
9x
2
– 6x + 1 and 9x
2
– 6x + 1 = 0,
How do they differ?
Now you will go further solving such expressions.
Solving Quadratic Equations
Activity 3
a. Describe a quadratic equation.
b. Write an example of a quadratic equation.
c. Solve the equation you came up with.
Report your work to the class.
Denition;
Quadratic equations are equations of the form ax
2
+ bx +c = 0.
There are different methods for solving quadratic equations.
Can you mention them? You are going to look at some of them.
a. The factor method.
How do you factorise these quadratic expressions? Now look
at these examples.
Example 3
Solve the following;
(a) x
2
+ x – 6 = 0
Solution;
Factorise the left-hand side as before
10
11
i.e. x
2
+ x – 6 = (x + 3)(x – 2) …. These are the factors
∴ (x + 3) (x – 2) =0
either x + 3 = 0 or x – 2 = 0
∴x = –3 or x = 2
(b) 3x
2
– 11x (– 20) = 0
Solution;
Multiply 3x
2
x(–20) = –60x
2
Find two factors of –60x
2
Adding give –11x and multiplying them give –60x
2
These are –15x and 4x
3x
2
– 15x + 4x – 20 = 0
3x(x – 5) +4x(x – 5) = 0 .................... factorise
(3x + 4) (x – 5) = 0 ............................. factor out (x – 5)
either 3x + 4 = 0 or x – 5 = 0
∴ x = –
4
/
3
or x = 5
Exercise 1c
Solve the following quadratic equations by factorisation
1. f
2
+ 3f + 2 = 0
2. s
2
+11s + 18 = 0
3. x
2
+7x + 6 = 0
4. r
2
+ 16r + 15 = 0
5. x
2
- 8x + 12 = 0
6. t
2
+ 3t – 10 = 0
10
11
7. p
2
–2p – 15 = 0
8. x
2
– 3x – 54 = 0
9. t
2
+ 12t + 27 = 0
10. m
2
– 3m – 10 = 0
11. x
2
+ 4x – 32 = 0
12. 2x
2
–5x + 2 = 0
13. 2n
2
− 10n + 12 = 0
14. 3d
2
+ 5d – 12 = 0
15. 35 + 30d −5d
2
You will then turn to the other method.
b. Completing the square of a quadratic expression
Activity 4
1. What do you call an expression such as (x + 1)
2
?
2. Make x
2
+ 8x + 16 in the form as above and hence solve
(x+4)
2
=0
3. Brainstorm on other quadratic expressions which can be
solved in the same way.
Dicuss your nding to the class.
The following are examples of the square terms: x
2
, (2x)
2
,
(x – 2)
2
,(x + 3)
2
etc.
Recall also that x
2
=16
4
16
±=
±=
x
x
So
25 3 25 )3(
2
±=−∴=− xx
2
or
8
53
−==
±=
xx
x
12
13
An expression x
2
– 4x + 4 can be factorised as (x – 2) (x – 2) =
(x – 2)
2
, a square.
Not all quadratic expressions can factorise into squares.
Expressions such as (x – 2)
2
, (x + 3)
2
are called Perfect
squares
You are going to learn a method of Completing the square of
any quadratic expression.
Example 4:
Completing squares
Solve the following quadratic equations by completing the
square.
a) x
2
+ 6x + 8 = 0
Divide the co-efcient of x by 2, add to x and square the
result.
i.e. (x + 3)
2
, if we expand this we get x
2
+ 6x + 9, we must
take away 1
i.e.
01
)
3
(
86
22
=−+=++ xxx
1
)
3
(
2
=+∴ x
13
±=+∴ x
13
±=+x
4or
2
−=−=∴ xx
b) 2x
2
–10x + 9 = 0
The rst step is to divide through by the co-efcient of x
2
, in
this case 2 i.e. x
2
– 5x + 4½ = 0
then x
2 –
5x = –4½ (taking – 4½ the other side)
x
2
–5x +
()
2
5.2
=
()
2
5.23.5
+−
(completing the square)
i.e
()
5.24.5
)
5.2
(
2
2
+−=−x
12
13
= − 4.5 + 6.25
75.1
5.2
±=−∴ x
(Removing the root)
1.75
−
2.5
or
75.1
5.2
=
+=
x
x
(Simplifying the RHS)
To get the value of
75.1
quickly, you need to use a calculator.
Some of the questions in the exercise below require the use of a
calculator.
Oral exercise
What must be added to the following expressions to make them
into a perfect square? Factorise the results
a. u
2
−u
b. d
2
– 6d
c. x
2
+10xy
d. a
2
– 6ad
e. y
2
– 3y
Exercise 1d
Solve the following quadratic equations by completing the
square. In cases where answers cannot be given in exact form,
give your answer correct to 2 d.p.
1. b
2
+ 4b + 3 = 0
2. c
2
– 2c + 1 = 0
3. t
2
– 14t + 48 = 0
4. e
2
– 6e – 16 = 0
5. x
2
– x – 6 =0
6. n
2
– n – 13 = 0
14
15
7. r
2
– 4r – 11 = 0
8. v
2
+v – 18 = 0
9. x
2
+ 2x – 7 = 0
10. 2h
2
– 3h – 4 = 0
11. 3x
2
– 4x –2 = 0
12. 4z
2
+ 2z – 5 = 0
13. –2u
2
– 5u +2 = 0
14. 5t
2
– 8t–1 = 0
15. –7x
2
–x +15 = 0
Having solved those above, can you also solve x
2
– x – 5 = 0,
what do you notice? How can such equations be solved? Do the
Activity 5 below.
The Quadratic formula
Activity 5:
A general quadratic equation is of formax
2
+ bx + c = 0, discuss
with a friend and solve for x just like example 4b ie
2x
2
–10x +9 = 0 above.
Compare with one given below.
Deriving quadratic formula
ax
2
+ bx + c = 0 then ax
2
+ bx = –c (taking c the other side)
a
c
a
bx
x
2
−=+
(dividing by a on both sides)
(add both sides)
2
2
4
a
b
+
−
=
a
c
14
15
a
c
x
−
+=
+
4
a
b
2
a
b
2
2
2
2
2
2
4
a
4
ac
b
2
−
=
+
a
b
x
(Simplifying the RHS)
a
acb
a
b
x
4
4
2
2
−
±=+
(Taking the square root on both
sides)
2
2
4
4
−
2
a
b
−
a
acb
x ±=
a
acb
x
2
4
2
a
b
−
2
−
±=
a
acbb
x
2
4
2
−±−
=
This is called the quadratic formula and it is used to solve
quadratic equations.
Note, b
2
– 4ac under the root is called a discriminant. It
indicates the number of solutions the quadratic expression has.
Example 5
Solve the following quadratic equation using the quadratic
formula.
x
2
+ 2x – 15 = 0
Here a = 1, b = 2 and c = –15
∴ Using the quadratic formula
12
)
15
(
142
2
2
×
−××−
±−=x
2
604
2
+
±−=x
2
64
2
±−=x
16
17
2
82
±−
=x
x = 3 or x = –5
Exercise 1e
Solve the following quadratic equations using quadratic formula.
In some cases, you need to use a calculator or log tables. Give
your answers to 2 d.p.
1. a
2
+ 7a + 12 = 0
2. x
2
– 2x – 24 = 0
3. p
2
+ 6p – 40 = 0
4. y
2
– y –6= 0
5. x
2
+ 2x – 8 = 0
6. e
2
+ e – 9 = 0
7. x
2
– 3x – 3 = 0
8. t
2
– 2t – 2 = 0
9. x
2
–x – 5 = 0
10.x
2
+ 2x – 7 = 0
11.5c
2
–8c +1 = 0
12 3z
2
–4z –2 =0
13. x
2
–6 = 0
14. x
2
–8 = 0
16
17
Formulating quadratic equations when roots are given
Activity 6a:
A car at times needs to reverse, so too in quadratic equations.
If a quadratic equation has roots x = 3 and x = −5, come up with
its equation. Share your work with neighbour. Did you solve as
those given below?
Example 6:
Formulating quadratic equations
Formulate the equation whose roots are:
a) x = 2 and x = –3
b) x = 5 and x = ½
a) If x = 2 then x – 2 was factor and
If x= –3 then x + 3 was factor
∴ the equation is (x – 2) (x + 3) = 0
i.e. x
2
+ x – 6 = 0
b) x – 5 and x – ½ were the factor so (x – 5) (x – ½) = 0 is the
equation
x
2
– 5½x +
5
/
2
= 0 …multiply by 2 throughout to get rid of the
fractions
2x
2
–11x + 5 = 0
Exercise 1f
Find the quadratic equation whose roots are:
1. x = 3 and x = 1
2. x = –7 and x = 2
18
19
3. f = –3 and f = –2
4. x = 0 and x = 4
5. x = –3 and x = 1
Is it possible to have everyday life situation expressed as
quadratic expression and solve it? Do the activity below.
Formulating quadratic equations from Practical problems
Activity 6b:
a. Get a piece of paper or card board which is rectangular.
b. Using a piece of paper or string equal to the width (y) of the
rectangular paper.
c. Place the piece of paper or string along the length of the
rectangle and mark where it ends.
d. Using a ruler, measure the remaining part of the length of
rectangle and record its reading.
e. Derive an expression for nding the area of the rectangle.
In most cases, you have to formulate quadratic equations from
everydaylife situations and solve them to nd the values of the
unknown like the one above.
Example 7:
Formulating quadratic equations
a. Benjamin is x years old and his sister Susan is 5 years
younger. If the product of their ages is 36, form an equation
in x and solve it to nd Benjamin’s and Susan’s age.
Benjamin is x years old Susan is (x – 5) years old and
x(x – 5) = 36 is the product of their ages
18
19
x
2
– 5x = 36
∴ x
2
– 5x – 36 = 0
Factorising we have (x – 9) (x + 4) = 0
∴ x = 9 or x = –4
–4 years does not make sense therefore x = 9
So Benjamin is 9 years old and Susan is 4 years old
b. When a number x is added to its square, the total is 12. Find
two possible values of x.
Let the number be x, the other is x
2
The equation is x + x
2
= 12
x
2
+ x –12 = 0 collecting the terms to one side two factors of
12 when added give +1 and multiplied give –12, 4 and –3 (x
+ 4) (x – 3) factorising
∴ x = –4 or x = 3.
Exercise 1g
In the following questions, formulate equations from the
information given and then solve to nd the unknown.
1. If the area of the rectangle above is 28cm
2
, calculate the
value of x and hence nd the length of the rectangle.
2. A rectangle is 5cm longer than it is wide. If its width is x
cm and its area is 66cm
2
form an equation in x and solve it.
Hence, nd the dimensions of the rectangle.
3. A right - angled triangle has a length of x cm and a height of
(x – 1)cm. If its area is 15cm
2
calculate the base length and
height.
4. The square of a number x is 16 more than six times the
number. Form an equation in x and solve it.
20
21
5. When ve times a number x is subtracted from the square of
the same number, the answer is 14. Form an equation in x
and solve it.
6. I think of a number x, if I square it and add to it the number
I rst thought of, the total is 56. Find the number I rst
thought of.
7. Grace has x
2
marbles. Duncan has x marbles. The sum of
their marbles is 90. Find the number of marbles each one
has.
8. Mercy is x years old and her sister is 5 years older. If the
product of their ages is 104, form an equation in x and solve
it to nd Mercy’s age.
9. A right-angled triangle ABC has <B = 90°, AB = x cm, BC =
2 cm longer than AB and AC is 4cm longer than AB.
a) Illustrate this information on a diagram.
b) Using this information show that x
2
– 4x – 12 = 0
c) Solve the above equation and nd the length of each of the
three sides.
Unit summary
• Quadratic equation is the expressions of the form
ax
2
+ x + c. These fall under monomial, binomial and
trinomial. The unit factorisation of quadratic has looked at
expressions, solving quadratic equations by factorisation,
completing square as well using quadratic formula. The unit
has also looked at formulating quadratic equations from
real life problems. The next unit looks at circle geometry.
Unit review exercise
1. Factorise the following quadratic expressions:
a. n
2
– 12n + 36
b. x
2
–15x + 36
20
21
c. h
2
– 5h – 24
d. x
2
– 20x + 36
e. b
2
– 5b – 36
2. Solve the following quadratic equations
a. y
2
– 4 = 0
b. x
2
– 144 = 0
c. 16r
2
– 25 = 0
d. 4p
2
– 81 = 0
e. e
2
– 2e –63 = 0
1. In order to deal with problems of climate change, a certain
village established a rectangular forest whose diagonal was
120m. If the width is 16m less than length, nd the length
of the forest.
2. During the 2014 tripartite elections for Member of
Parliament Mr Chimuzu got 90 votes extra than Mr
Chitsinde. The winner Mrs Masamba got 3600 votes which
was the product of the votes of the rst two. Find the total
votes for Mr Chimuzu.
3. The number of HIV/AIDS patients in two consecutive years
increased by 5. If the product of the number of patients in
the two years was 50, nd the number of HIV/AIDS patient
in the rst year.
4. The oversized television at the left has a 60-inch diagonal.
The screen is 12 inches wider than it is high. Find the
dimensions of the screen.
Glossary
Quadratic expression: is an algebraic expression of the form
ax
2
+ bx + c, where a, b and c are constants and a ≠ 0
Monomial: is an algebraic expression containing one term eg
2x
2
.
Binomial is an expression containing two terms, for example
5x
2
+ 2x.
22
23
Trinomial is an algebraic expression containing three terms,
eg 2x
2
+ 6x + 7
Perfect square is an algebraic expression of the form (a + b)
2
References
Chanon J. B. & etl, New General Mathematics 3, a Modern Course
for Zimbabwe London: Longman.
Hau S. and Saiti F. (2010). Strides in Mathematics 3. Blantyre:
Longman Malawi.
22
23
Unit
2
IRRATIONAL NUMBERS
In your JCE Mathematics you
learnt about numbers. You learnt
about types of numbers such as
natural numbers, whole numbers,
and integers. In this unit you will
learn about irrational numbers.
You will learn how to recognise
irrational numbers from a given
set of numbers. You will also learn
about a special type of irrational
numbers called surds and how you
can simplify them.
Irrational numbers are important
as they assist in being exact with
certain measurements. For example,
irrational numbers are used in
expressing angles in trigonometry,
nding length, areas, and volumes of
objects.
Recognising rational and
irrational numbers
Numbers which can be expressed
exactly as fractions are known as
rational numbers. For example:
4
,
2
1
= 0.5,
4
3
= 0.75
However, there are certain numbers
which cannot be expressed as exact
fractions such as:
3
...73205
.1
=
and
7
= 2.645751311…. Pi or π.
Decimals, in these examples, extend
forever and are non- recurring
or repetitive. Most of these are
expressed as roots. They are also
known as surds.
From the statements above, you
can see that some numbers, which
contain a root sign, can be evaluated
exactly while others cannot.
Now you are going to look at
irrational numbers in details.
Activity1:
Recognising rational and irrational
numbers.
24
25
1. Dene rational and irrational numbers?
2. Give examples of irrational numbers.
3. Which of the following are irrational numbers?
(a) 3.142857…
(b) 0.75
(c)
5
(d) 5.252525…
Report your answers to the class.
Compare your work with the statements given below.
Rational and irrational numbers
• Rational numbers are numbers that that can be expressed
as ratio of two integers. For example ¾ = 0.75, and 5.2525...
are rational numbers.
• Irrational numbers are numbers that cannot be expressed
as a ratio of two integers.
In the activity above (a) 3.142857 and (c) √5 = 2.23606..are
irrational numbers. These numbers extend forever without
repetition or recurring.
Exercise 2a
1. Which of the following are rational and which are
irrational?
(a) 7, (b) √16, (c) 0.824, (d) 0.7, (e) √3, (f) √17 (g) √8, (h) √99,
(i) √121
2. Draw aright triangle ABC with angle ABC = 90
0
and
AB = BC =1cm . Find AC in surd form. Is the AC rational or
irrational number?
24
25
Surds
Many roots are irrational, for example; √3 = 1.732050… √21 =
4.5825... Irrational numbers of this kind are called surds. Thus
if the root of a number is irrational, then it is called a surds. In
other words; a surd is a root of irrational number.
Rules for surds
Activity 2:
Deriving the rules for surd;
In groups do this activity
By putting N = 16 and M = 9, nd which of the following pairs of
expressions are equal:
a.
NM
and
M
×
N
b.
N M
and
N
×
M
c.
M
×
M
and
M
2
d.
N
M
and
e. M
N
and
From your results, what rules can you come up about surds?
Report your nding to class.
You should have noted that each of the pairs is equal. Therefore
in general, here are the rules of surds.
a.
M
=
M
×
N
b.
N
M
=
c. M
N
=
NM
2
d.
M
×
M
=
M
2
= M
NM
2
M
N
N
M
N
26
27
c. M
N
+ P
N
= (M+P)
N
d. M
N
− P
N
= (M − P)
N
Simplifying surds
In simplication of surds, numbers under the square root sign
are reduced as much as possible. This is done by expressing the
numbers under the square root sign as a product of two numbers
in which one of them is a perfect square.
Activity 3:
Simplifying surds
In pairs, apply knowledge from activity above.
Simplify each of the following as far as possible by applying the
rules above.
(a)
12
(b)
20
(c)
32
Check your answers against your friends’ work.
Now look at the following examples and compare them with your
work above.
Example 1:
Simplifying surds
Simplify each of the following as far as possible.
(a)
162
(b)
3
6
3
Solutions
In each case, you must look for two factors for a given
26
27
number of which one should be a perfect square. I.e. √80 =
√16 × √5 and 16 is a perfect square.
(a) Solution;
162
=
81×2
........ simplify 162 = 81 × 2
=
81
×
2
...........… surd rule
= 9 ×
2
…................. nd root of 81
= 9
2
(b) Solution;
3
6
3
=
3
9 × 7
=
9
×
7
................... Surd rule
3
7
3
×
=
........................ Cancel out 3
=
7
Now do the following exercise
.
Exercise 2b
Simplify each of the following:
(1)
45
(2)
147
(3)
112
(4)
50
(5)
7
98
(6)
72
(7)
282
(8)
200
(9)
15
90
(10)
8
(11)
343
(12)
48
28
29
(13)
7500
(14)
9
18
Addition and subtraction of surds
You can subtract and add surds just like other ordinary
numbers.
Activity 4:
Addition of surds
Find the solution of the following;
(a) 2b + b
(b) 2√m
2
+ m
(c)
553
+
Discuss your ndings as a class and compare your answers with
the examples given below:
The knowledge above also applies when adding surds. Check the
following examples.
Example 2:
Simplifying surds
Simplify the following.
(a)
520
+
=
554
+×
=
554
+×
----------------- nd factors of 20 = 4 × 5
=
552
+
--------------------- this is the same as 2a + a
= 3
5
(b)
504185
+
=
2254 295
××
×+×
28
29
= 5
225429
××−××
since
MN
=
M
×
N
= 5
25423
×+×
............. taking root of 9 and 25.
=
220215
+
= 35
2
(a) 3
1275
−
= 3
34325
×−××
.......25 and 4 have roots
= 3
34325
×−××
............separating the roots
= 15
323
−
...................subtract just like 15b-2b
= 13
3
Exercise 2c
Simplify each of the following
(1)
17528263
+−
(6)
75312
+
(2)
72218200
−+
(7)
22318
−+
(3)
503218
+−
(8) 5
294246
+−
(4)
775511
−+
(9)
75248108273
−−+
(5)
504185
+
(10)
2
2
−
1
2
5
Multiplication of surds
When two or more surds multiply each surd is simplied as
far as possible and then numbers in front of square roots signs
multiply each other and then surds multiply with other surds.
Example 3
Simplify the following.
(a)
12
18
×
(b)
53
75
×
30
31
Solutions
Where possible look for two factors for a given number
under the root of which one should be a perfect square as
you did above and nd its root.
For example: 80
516
×
=
=4
5
and multiply as you do with
ab × ab.
(a)
12
18
×
34
×=
29
××
..............simplify
32
×=
23
××
............. nd root of 4 and 9
=
2323
××
.................... rule of surd
236
××=
66
=
(b)
20
27
×
54
×=
39
××
...............simplify
4
=
5
×
9
×
3
×
.............surd rule
52
×=
33
××
.................nd root of 4 and 9
532
××=
3
×
156
×=
156
=
Proceed as in a and b above
(c)
53
75
×
53
=
325
××
53
=
325
××
53
=
35
××
553
××=
3
×
1515
=
30
31
Exercise 2d
Simplify the following
a) √5 × √10
b) √2 × √6 × √3
c) √30 × √5
d) √12 × √3
e) √32 ×√12
f) √10 × 3√2 × √20
g) √5 × √24 × √30
h) (2√3)
3
i) (2√7)
2
j) √6 × √8 × √10 × √12
Express each of the following as the square root of a single
number, using rule number (c) for surds.
(1)
62
(2)
73
(3)
62
(4)
212
(5)
102
(6)
27
(7)
33
(8)
512
(9)
36
(10)
172
(11)
35
(12)
222
(13)
732
(14)
54
(15)
153
Division of surds
When surds divide, both the numerator and denominator should
be simplied as far as possible and then denominator should be
rationalised.
1.
2.
32
33
Example 4:
Simplifying surds
Simplify the following.
(a)
27
3
(b)
75
72
Solutions
Find two factors of a number under root as before
(a) Solution;
27
3
=
39
3
×
........... simplify 27 = 9 × 3
=
39
3
×
............. rule of surds
=
33
3
×
….......... nd root of 9 which is 3
=
3
1
…..................... cancel out 3
=
3
1
×
3
3
….............. multiply numerator and
denominator by √3
=
3
3
(b) Solution;
75
72
=
325
236
×
×
=
325
236
×
×
…........….. simplify by surd rule
=
35
26
………...…. nd root of 36 and 25
32
33
=
3
3
35
26
×
….......... Multiply numerator and
denominator by √3
=
15
66
……………..divide 6 and 15 by 3
=
5
62
Exercise 2e
Simplify the following;
1.
32
8
2.
123
182
3.
4.
5.
6.
7.
8.
9.
10.
11.
34
35
12.
Conjugate surds
Activity 5:
Expanding conjugate surds
(a) Factorise (a
2
− b
2
)
(b) What term is used to describe a
2
− b
2
?
(c) Expand (x + y) (x − y)
Relate your answers to the explanation given below.
Conjugate surds are surds of the form
( )
78 −
and
( )
78 −
,
for example
( ) ( )
.5252 −+ and
when multiplying conjugate
surds, they are expanded the same way as you do with factors of
a difference of two squares.
Example 5:
Conjugates
Conjugate and expand the following surds
(a)
35 +
(b)
35 +
(c)
35 +
Solutions
First come up with a conjugate of
35 +
in (a) which is
35 +
34
35
(a)
expand as you do with (a + b) (a – b)
=√2 (√2 – √5) + √5 (√2 – √5) …......… as a(a – b)+ b(a – b)
=
55252522 ×−×+×−×
= 2 + 0 – 5
= –3
(b)
( )( )
2323 −+
do as above and this leads to
=
()
410105
2
−−+
, since
1010 −+
gives zero
= 5 – 2
= 3
(c)
=
( )
4663
2
−+−
, since
66 +−
= 0
= 3 – 2
= 1
Exercise 2f
a. Find the conjugates and expand each of the following surds:
(1)
( )
ba +
(2)
( )
ba +
(3)
( )
78 −
(4)
35 +
(5)
( )
ba +
(6)
(
)
411 −
(7)
(
)
1113 +
(8)
( )
ba +
(9) 3 +
8
(10)
()
2343 −
36
37
b. Simplify the following without using a calculator.
1. (3√2 − √3) (3√2 + √3)
2. (2−√7) 2(2 +√7)
3. 2/√2 + 6/√2
Rationalising surd denominators
When the denominator of a fraction is a surd, it is very
necessary to remove the surd from the denominator. This is
known as rationalizing the denominator i.e. the irrational
denominator is changed to a rational denominator. The process
involves multiplying the number by 1 ie , which takes different
forms depending on the denominator in question.
Activity 6:
Rationalising denominator
Try to rationalise the denominator by multiplying numerator
and denominator by the denominator in each of the following:
(a)
5
2
(b)
3
54
Present your work to the whole class or show your friends and
compare with the example given below.
Example 6
Rationalise the denominator of the following;
a.
5
2
In question (a), the 1 to be multiplied is
4
53
+y
5
52
=
5
5
5
2
×=
36
37
1
3
6
×
5
52
=
b.
3
6
In question (b), the 1 to be multiplied is
4
53
+y
1
3
6
×
=
3
3
3
6
×
=
313
7
−
3
18
=
This can be simplied further to give √2, work out!
Note: In general, if a fraction is in the form of
5
7
, then multiply
both the numerator and the denominator of the fraction by
8
.
c.
53
20
Solution
Do as in above ,multiply by
4
53
+y
53
20
× 1
=
53
20
×
4
53
+y
38
39
=
53
520
×
×
=
15
520
=
3
54
Sometimes, the fraction is in the form
ba +
1
. In this case,
you have to multiply both the numerator and denominator by
( )
ba +
Similarly, if the fraction is in the form
3
y
zx
, then multiply
the numerator and denominator by
( )
ba +
.
d.
15
2
+
Solution
( )
32
1
+
×
( )
( )
32
32
−
−
…...................… from statement above
=
932324
32
−+−
−
….................... Expanding
=
34
32
−
−
=
15
2
+
= 2 −
8
38
39
Exercise 2g
Rationalise the denominator of the following
(1)
5
7
(2)
5
7
(3)
5
7
(4)
5
7
(5)
5
7
(6)
5
7
(7)
5
7
(8)
15
2
+
(9)
15
2
+
(10)
15
2
+
(11)
15
2
+
(12)
4
5
3
+y
(13)
35
35
+
−
(14)
713
713
+
−
(15)
313
7
−
Real life problems of irrational numbers
In most cases real life problems of irrational numbers involve
nding lengths, areas, and volumes of given objects. Most of
such measurements are actually irrational numbers. The use of
which is irrational number is more common in nding volume
and areas of certain objects.
Exercise 2h
1. The base length of an isosceles triangular truss of a house is
14m. The height is 2m. Find the approximate length of iron
sheets that can be bought assuming that they are equal to
the third side.
2. The two sides of a right angled triangle are
7
−3)and
(3 – 3) cm. Find the length of hypotenuse.
3. Find the perimeter of the a triangle whose sides are 2√48
cm, 3√75 and 4√147 cm. The answer should be in simplied
40
41
surd form.
4. A right angled ABC with AB = 5, BC = 7cm. Find the length
of hypotenuse side AC in surd form.
Unit summary
• In this unit you have learnt about rational numbers which
are numbers that can be expressed as exact fractions. You
have also learnt about irrational numbers which cannot
be expressed as exact fractions. Most irrational numbers
are under root and as such are called surds.You have
learnt how you can simplify surds in addition, subtraction,
multiplication and division. In the next unit you are going to
learn about circle geometry.
Unit review exercise
1. Without using a calculator, simplify the following
a
(
24
– 6
)
2
6
b √125 + √5 – √45
c (3√5 – √2)/(2√5 + 3√2)
d
35 – 2
25+3 2
e √99 – √44 – √11
f 3√28 – 5√63 + 4√252
g 2√150 – √96 – 2√24
g (8√2)/(√98 – 3√2)
i 1/√2 – √2/3
j (√2 + √3)(√8 – √12)
2. Given that √2 = 1.414 and root √3 = 1.732 evaluate the
40
41
following;
(a) 1/√3
(b) 3/√3
(c) 10/(√2)
(d) 6/(√3)
(e) 2/(√2)
3. Rationalise the denominator of the following
(a) 1/(5 + √2)
(b) 1/(4 – √3)
(c) (5 + √3)/(√7 + √5)
(d) √(3 + 1)/√(3 – 1)
(e) (3 + √5)/(√3 – √2)
Glossary
Rational numbers are numbers which can be expressed as
exact fractions or ratios.
Irrational numbers are numbers which cannot be written as
exact fractions. The decimals extend without end and without
recurring.
Surds are square roots of irrational numbers.
Reference
Hau and Saiti F. (2010). Strides in Mathematics 3. Blantyre:
Longman Malawi.
Elaine R., McAdams P, Hiddleson P ., (2013), CHANCO Teach
yourself series, Mathematics, Second Edition, Chancellor
College Publication, Zomba.
42
43
Hardwood Clarke and Norton F. G. J., (1984) Seventh Edition,
Oxford, Heinman Educational Book Ltd, London
Channon J. B. et al (1997). New General Mathematics 3, A Modern
Course for Zimbabwe.Longman London.
Gunsaru et al, Secondary Mathematics Book 3, Dzuka Publising
Company, Blantyre.
42
43
Unit
3
CIRCLE GEOMETRY (CHORD
PROPERTIES)
In unit 2 you learnt about irrational
numbers. In this unit, you will look
at the chord properties of a circle.
Specically, you will describe chord
properties of a circle and apply
chord properties to solve problems.
Chord properties of a chord
A circle is an important shape in the
eld of geometry. A circle is a shape
with all points the same distance
from its centre. A circle is named by
its centre. Before going in detail, do
activity 1 below.
Activity 1:
Identifying parts of a circle
In pairs,
i. Dene radius.
ii. Using a pair of campus, draw
circle with centre O and label
the following parts, diameter,
radius, a chord, an arc,
circumference and segment.
iii. Brainstorm on the labeled
parts.
Compare your work with the drawing
below.
Parts of a circle Now look at gure
3.1a below which is showing some
parts of the circle.
Figure 3.1a
E
A
44
45
The distance around a circle is called the circumference of a
circle. Figure 3.1a is a circle with centre O.
A chord is a line segment that joins two points on a circle. Line
EF is a chord.
A diameter is a chord that passes through the centre of the
circle. Line AB is diameter.
A radius is a line segment that has the centre and a point on the
circle as end points. In gure3.1a, AO, OB and OC are radii
Arcs of a circle A secant is a line containing a chord. If a
secant intersects a circle in two distinct points, A, B for example
in the diagram below, then the circle is divided into two sets of
points.
B
A
B
Figure 3.1b
APB is known as the minor arc while AQB is known as the
major arc.
A sector of a circle is a region consisting of the union of an arc,
the radii to the end points of the arc, and the interior points
enclosed by the arc and the radii.
•
Sector
Segment
Figure 3.1c
44
45
An area bounded by chord and an arc is called a segment of a
circle.
There might be major and minor segment. Identify these in the
given circle.
The area bounded by chords and an arc is called a sector.
Chord property of circle
A chord is a straight line that join any two points of the on its
circumference
A chord which passes through the centre of a circle is called a
diameter.
(a) The straight line drawn from the centre of a circle to the
mid – point of a chord is perpendicular to that chord.
Conversely if a line is drawn from the centre of the circle
perpendicular to a chord, it bisects the chord.
(b) If chords are equal in length then they are the same
distance from the centre of a circle. Conversely if chords are
the same distance from the centre of a circle then they are
equal in length.
(c) If two chords of a circle are parallel, then the perpendicular
bisector of one is also perpendicular bisector of the other.
Chord theorems: In your JCE Mathematics , you learnt about
theorems. How did you dene a theorem?
Now look at the following gure
O
M
A
B
Figure 3.2
46
47
In gure 3.2, AB is a chord of the circle with Centre O. OM is
perpendicular to a chord AB. Measure AM and BM. What do you
notice?
Theorem 1: The line from the centre of the circle perpendicular
to a chord bisects the chord.
You should show the effect of OM being perpendicular to AB.
A
M
O
B
Figure 3.3
Given : Circle centre O, AB is a chord, OMAB.
To Prove : AM = MB.
Construction : Join OA, OB
Proof : In
OAM and
OBM
OA = OB (radii)
∠
OMA =
∠
OMB = 90
0
(given)
OM is common.
∴
OAM
≅
OBM (RHS)
∴
AM = MB
Conversely, the line from the centre of the circle bisecting the
chord is perpendicular to the chord.
46
47
•
M
O
A
B
Figure 3.4
Given : Circle, centre O, AB a chord AM = MB
To Prove :
∠
OMA =
∠
OMB = 90
0
Construction : Join OA, OB
Proof : OMA and OMB
OA = OB (radii)
AM = BM (given)
OM is common
∴
OMA
≅
OMB (SSS)
∴
∠
OMA =
∠
OMB
But AMB is straight line
∴
∠
OMA = OMB = 90
0
(
∠
s on a straight line)
∆
∆
48
49
Equal chords: A circle can have several chords as shown in the
following gure 3.5 below.
A
B
D
•
C
H
E
G
F
O
Figure 3.5
In this gure CD, EF and GH are all chords. In a diagram
with equal chords, you want to nd out what happens to their
distances from the centre of the circle.
Activity 2:
Identifying distance of equal chords from the centre.
In pairs:
a. Draw a circle with centre O and radius of 5cm.
b. From the centre, draw a line of 3cm to any point E and
another to any point F within the circle in opposite direction
of the rst line.
c. Draw a chord from a point A on one side of the circle passing
through E to point B on the other side of the circle.
d. Produce another line from point C on the circle passing
through F to point D on the other side of the circle.
e. Measure the length of the chord AB and CD.
f. What conclusion can you draw?
Report your ndings to the whole class.
Now consider the following theorem.
48
49
Theorem 2: Equal chords are equidistant from the centre of the
circle.
•
O
A
B
D
C
Figure 3.6
Figure 3.6 is a circle with centre O, and AB is a chord that is
equal to chord CD. You want to show that the shortest distances
(the perpendicular distances) from centre O to the chords, are
equal.
A B
DC
•
Q
P
O
Figure 3.7
Given : Circle centre O, AB = CD
To Prove : OP = OQ
Construction : Join OB, OD
Proof : Since OP
⊥
AB
∴
AP = PB (since line from centre bisect the chord as proved
already)
i.e. PB =
3
4
AB
And OQ
⊥
CD
50
51
CQ = QD (as above)
i.e. QD =
3
4
CD
But AB = CD (given)
PB = QD.
In OPB and OQD
PB = QD (proved)
∠
OPB =
∠
OQD = 90
0
(given)
OB = OD (radii)
∴
OPB
≅
OQD (RHS)
OP = OQ. (Corresponding sides)
Conversely, it can be proved that chords that are equidistant
from the centre of the circle are equal.
•
M
N
B
O
A
C
D
Figure 3.8
Given : Circle centre O, OM = ON
To Prove : AB = CD.
Construction : Join OB, OD
Proof : OMB and OND
OM = ON (given)
∠
OMB =
∠
OND = 90
0
(given)
50
51
OB = OD (radii)
∴
∆
OMB
≅
∆
OND (RHS)
∴
MB = ND
But OM
⊥
AB
MB =
3
4
AB
Similarly, ON
⊥
CD
∴
ND =
3
4
CD
But MB = ND (proved)
∴
AB = CD
Now do the following exercise
Exercise 3a
1. Given that a circle with centre O and any two points P and
Q on the circumference of the circle such that PQ is not
a diameter, E being the midpoint of PQ, prove that OE is
perpendicular to PQ.
2. Prove that if two chords bisect each other, then the chords
are diameters.
3. Prove that the perpendicular from the centre of a circle to a
chord bisects the chord.
For questions 4 and 5, refer to the diagram Figure 3.9
below:
52
53
C
K
O
A
H
B
Figure 3.9
4. OH, OK are the perpendiculars from the centre O of a circle
to two chords AB, CD, if AB = CD, prove that OH = OK.
(i) Explain why AH = CK
(ii) Prove triangle OHA ≡ triangle OKC
5. OH, OK are perpendiculars from the centre O of a circle to
two chords AB, CD. If OH = OK, prove that AB = CD.
(i) Prove
OHA
OKC
(ii) Explain why AB = 2AH and complete the proof.
6. If the chord AB, length l cm, of a circle, radius r cm, is at a
distance p cm from the centre, nd l in terms of r and p
7. AB and CD are two equal chords of a circle; M, N are their
mid – points. Prove that MN makes equal angles with AB
and CD.
8. Two chords of a circle bisect each other at K. prove that
K is the centre of the circle. Hint; if possible, join K to the
centre.
Application of chord properties to solve problems
Using the knowledge from above, you can now nd radius,
length of the chord and distance from the centre to a chord.
Calculating the length of the chord given radius and
distance from the centre
D
52
53
Activity 3
Calculating the length of the chord given radius and
distance from the centre
In pairs, use the information given;
The gure 3.10 is circle with centre O and radius of 5cm and
OE = 3cm. calculate the length of the chord AB.
A
B
O
E
Figure 3.10
Present your work to class and compare your work with the
given example.
Calculating the distance from the centre given the length of
chord and radius
Example
Two parallel chords of a circle are of length 16cm and 12cm. If
the radius of the circle of the circle is 10cm, what are the two
possible perpendicular distances between the chords?
Solution
Draw the two chord in this circle; let AB = 16cm and CD = 12cm
centre O, radius = 10cm
B
D
A
C
O
E
F
Figure 3.11
54
55
In
AOE, using Pythagoras theorem
OE
2
= AO
2
– AE
2
= 10
2
– 8
2
since AE = BE as OE bisect chord AB
= 100 – 64
= 36
OE = √36 = 6cm
Similarly in
COF
OF
2
= OC
2
– CF
2
= 10
2
− 6
2
= 100 – 36
= 64
OF = √64 = 8cm
The distances from the centre are OE = 6cm to chord AB and
OF = 8cm to chord CD.
Calculating the radius given length of a chord and distance
from the centre
Example
A chord 4.2cm long is 2.8cm from the centre of a circle. Calculate
the radius of the circle.
Solution: draw diagram like the one below
M
N
O
P
Figure 3. 12
54
55
In triangle OPN in gure 3.12
ON
2
= OP
2
+ PN
2
= 4.2
2
+ 2.8
2
= 17.64 + 7.84
ON = √25.28
ON = 5.048cm
Exercise 3b
1. Use gure3.13 to answer the questions given below.
•
B
O
A
Y
M
x
Figure 3.13
1. Using Pythagoras theorem in gure 3.13, calculate the
following;
a) OM if AB = 10cm and OA = 7cm
b) AM if OM = 3cm and OA = 8cm
c) OA if AB = 12cm and OM = 6cm
2. Two parallel chords of lengths 6cm and 8cm are drawn in
a 5cm radius circle. Calculate the two possible distances
between them.
3. A circle has a radius of 10cm. Use Pythagoras Theorem to
nd the length of a chord of the circle that is 6 cm from the
centre of the circle. How long is a chord of the same circle
that is 8cm from the centre?
56
57
4. The gure below shows a circle ACB centre O. OM = 3y cm,
MC = 2y cm and angle OMB = 90
0
.
O
M
B
A
C
Figure 3.14
Find AB in terms of y
5. Chord is 6cm away from the centre of the circle. If the chord
is 16cm long, calculate the radius of the circle.
6. A chord 20cm long is 20cm from the centre of a circle.
Calculate the length of the chord which is 14cm from the
centre.
7. In a circle of radius 2.5cm the lengths of two parallel chords
are 1.4cm and 3cm. nd the distance between the chords;
(a) If they are on opposite sides of the centre
(b) If they are the same side of the centre.
8. A chord of a circle of radius7cm is at a distance of 4cm from
the centre. Calculate the length of the chord.
9. AB, CD are parallel chords of a circle, 3cm apart, on same
side of the centre O; AB = 4cm, CD = 10cm, nd OA. Hint;
draw ONM perpendicular to AB to meet AB, CD in M, N;
let ON = x cm.
10. A chord of length 10cm is at a distance of 12cm from the
centre of the circle. Find the radius.
Unit summary
• In this unit you have learnt about chord properties of a
circle and how to apply them to solving problems. You
56
57
have also learnt on how to nd radius, length of a chord
and distance from the centre. In the next unit, you will
learn about algebraic fractions with linear or quadratic
denominators.
Unit review exercise
1. A chord 7cm long is drawn in a circle of radius 3.7cm.
Calculate the distance of the chord from the centre of the
circle.
2. A chord 6.6cm long is 5.6cm from the centre of the circle.
Find the radius of the circle. Find also the length of a chord
which is 6.3cm from the centre of the circle.
3. Figure 3.15 shows a circle AYBX centre O. XY cuts AB at C
such that angle ACY = 90
0
.
B
X
Y
A
C
O
Figure 3.15
If OC = 8cm and CY = 9cm, calculate the length of AB.
4. Prove that if two chords of a circle are equal, then they are
equidistant from the centre.
5. Two equal chords intersect inside a circle. Prove that the
line joining their point of intersection to the centre of the
circle bisects the angle between the chords.
6. XY is a diameter of a circle, and XZ is a chord. If O is the
centre and OD is the perpendicular from O to XZ, prove
that XY = 2OD.
58
59
Glossary
A chord is a line segment that joins two points on a circle.
A diameter is a chord that passes through the centre of the
circle.
A secant is a line containing a chord
References
Hau and Saiti F. (2010). Strides in Mathematics 3. Blantyre:
Longman Malawi.
Channon J. B. and et al (1997). New General Mathematics 3, A
Modern Course for Zimbabwe. London,Longman.
Durell C. V . Volume 2. (1990) Certicate Mathematics. 2
nd
Edition. Bell and Hyman Ltd. London.
Gunsaru et al, Secondary Mathematics Book 3. Dzuka
Publishing Company. Blantyre. Malawi.
58
59
Unit
4
ALGEBRAIC FRACTIONS
WITH LINEAR
OR QUADRATIC
DENOMINATORS
You learnt about algebraic fractions
in your JCE Mathematics with
respect to LCM and HCF. In this unit,
you are going to learn about the four
operations: addition, subtraction,
multiplication and division involving
algebraic fractions . You are also going
to express fractions in their simplest
form.
Before all this, you must know how to
simplify algebraic fractions.
Algebraic fraction help to stimulate
critical thinking and reasoning ability
as mostly it uses the brain since
some fractions cannot be done by the
calculator. Skills acquired assist in
solving algebra and other problems in
mathematics.
Expressing algebraic
fractions to their lowest
terms
An algebraic fraction is a fraction
whose numerator and denominator
are algebraic expressions. In other
words it uses variables (unknowns) in
the numerator and denominator
Can you write some examples
of algebraic fractions? Here are
examples of algebraic fraction as a
reminder.
2
a
,
2
2
+a
a
,
2
4
2
+
−
x
x
,
9
65
2
2
−
++
x
xx
Can you
come up with more other example.
The rules for simplifying algebraic
fractions are the same as those for
numerical fractions. Now do the
activity below.
Expressing algebraic
fractions to their lowest
terms
Activity 1:
Reducing algebraic fractions to
their lowest terms.
In pairs, express the following
algebraic fraction in their simplest
form;
60
61
Simplify the following:
1.
3
8
12
3
y
y
2.
36
5
4
3.
1
3
2
3
Present your work to class. Discuss the examples given below.
Example 1:
Expressing into lowest terms
Express the expression to its lowest term,
48
4 2
18
2 6
Solution
Find factors that can go into the denominator and the
numerator and cancel out.
3
8
62
24
18
48
yx
yx
=
4
2
3
8
y
x
and we cannot simplify any further.
If you have the same variables on denominator and numerator,
subtract the powers. I.e. x
4−2
.
Exercise 4a
Simplify the following
1. (a)
4
6
x
x
(b)
3
12
y
y
(c)
p
p
6
(d)
4
5
m
m
2. (a)
22
64
yx
yx
(b)
cab
cba
2
422
(c)
rpp
rpq
32
22
(d)
yz
xy
2
4
+
60
61
(e)
p
pq
5
15
2
(f)
22
2
24
16
nm
nm
Addition and Subtraction of Algebraic Fractions
When adding or subtracting algebraic fractions, follow the same
procedure as in arithmetic fractions:
Activity 2:
Addition and subtraction
Simplify the following
(a)
⅜ + ⅛
(b)
1
2
+
2
3
(c)
3
4
1
3
(d)
1
4
8
Compare your work with the other members of the class.
In simplifying these, did you follow the following steps?
(1) Find the LCM of the denominators.
(2) Express each fraction with the common denominator (divide
the LCM by individual denominators and multiply by
respective numerators).
(3) Add or subtract the fractions.
Now look at the following examples.
Example 2:
Simplifying fractions
62
63
Simplify
(a)
3
8
12
3
y
y
Here, the denominators are x + 1 and x − 3 and their LCM is
(x + 2) (x − 3)
∴
(x(x − 3) − {(x + 2)(x − 2)}
(x + 2)(x − 3)
Divide LCM each denominator
then multiply
=
(x
2
− 3x − {x
2
− 4})
(x + 2) (x − 3)
expand by removing the denominator
=
(x
2
− 3x − x
2
+ 4
(x + 2) (x − 3)
=
4 − 3x
(x + 2) (x − 3 −)
(b)
5
32
3
53 −
+
− xx
The LCM of 3 and 5 is 15
∴
5
32
3
53 −
+
− xx
=
()()
15
323535 −+− xx
...… divide LCM by 3 and 5
=
15
962515 −+− xx
……... expand and add like terms together.
=
15
3421 −x
(c)
2
2
1
3
−
+
+ xx
The LCM of x + 1 and x − 2 is (x + 1) (x −2)
62
63
∴
2
2
1
3
−
+
+ xx
=
()()
()()
21
1223
−+
++−
xx
xx
……. Divide the LCM with denominators
=
()()
21
2263
−+
++−
xx
xx
……… expand and add like terms together
=
()()
21
45
−+
−
xx
x
Sometimes, there is need for factorising in order to nd the LCM
of the denominators.
(c)
2
5
4
3
2
−
−
− xx
x
2
− 4 is not as simple as x-2, it factorises to (x + 2) (x − 2)
∴
2
5
4
3
2
−
−
− xx
=
()()
2
5
22
3
−
−
−+ xxx
The LCM is (x + 2) (x − 2)
∴
()
()()
22
253
2
5
4
3
2
−+
+−
=
−
−
− xx
x
xx
=
()()
22
1053
−+
−−
xx
x
=
()()
22
75
−+
−−
xx
x
Now do the exercise given below.
64
65
Exerxise 4b
Simplify the following algebraic fractions.
1. (a)
54
aa
+
(b)
43
xx
−
(c)
654
xxx
++
(d)
12
7
8
5 ap
−
(e)
64
3 xx
−
(f)
432
zyx
−+
2. (a)
pp 3
2
5
3
−
(b)
yyy 5
4
3
53
+−
(c)
2
2
53
xx
+
(d)
s
r
s
r
s
r
21
5
14
2
7
3
−+
(e)
p
n
p
n
p
m
30
11
5
4
15
3
−+
3. (a)
4
12
3
23 +
+
+ xx
(b)
2
3
4
53 baba −
−
+
(c)
5
2
4
2
+
−x
(d)
6
7
3
3 −
−
− xx
(e)
2
73
6
53 nmnm +
−
−
(f)
(2a + 1) + (3b − 2 ) − 2
a b
4. (a)
3
3
4
2
+
+
+ xx
(b)
5
3
3
5
−
−
+ pp
(c)
3
2
2
3
+
+
+ yy
(d)
3
2
1
3
+
−
− mm
(e)
25
7
23
5
+
−
− xx
64
65
5. (a)
62
2
3
3
−
+
+ xx
(b)
1
2
1
4
2
−
−
− xx
(c)
2
4
4
3
2
−
+
−
x
x
x
x
(d)
12
3
9
2
22
−+
+
− xyy
(e)
152
32
209
43
22
−+
−
−
++
+
xx
x
xx
x
(f)
+1
+3
1
2
Multiplication of algebraic fractions
The simplication involves multiplying two algebraic fractions.
As with ordinary arithmetic fractions, numerators can be
multiplied together as can denominators, in order to form a
single fraction. Remember the following laws of indices
1. a
m
x a
n
= a
m+n
2. a
m
÷ a
n
= a
m−n
3. (a
m
)
n
= a
mn
Activity 3:
Simplifying algebraic fractions
Individually, simplify (a)
2
3
×
1
6
(b)
32
ba
×
Discuss your answer with a friend. Compare the way of solving
with the examples below;
Example 3:
Simplifying fractions
Simplify the following;
2
3
2
6
9
8
4
3
y
p
p
y
×
+
66
67
Factors, which are common to both numerator and denominator,
may be cancelled. It is important to realise that this cancelling
means dividing the numerator and denominator by the same
quantity.
.i.e.
2
3
2
6
9
8
4
3
y
p
p
y
×
3
2
4
py
=
Exercise 4c
1. (a)
34
yx
×
(b)
3
2 q
p
×
(c)
r
q
q
p
×
(d)
xy
s
r
xy
×
(e)
16
8 x
x
×
2. (a)
x
py
py
yx 4
3
2
×
(b)
p
q
q
p
3
5
2
3
2
×
(c)
22
3
6
a
b
b
a
×
(d)
bc
cd
b
ad
c
ab
4
8
2
6
2
××
(e)
2
3
2
2
2
2
3
10
5
6
3
2
y
c
zy
a
ac
yz
××
Division of algebraic fractions
Division and multiplication of algebraic fraction goes hand in
with multiplication
Activity 4:
Dividing of algebraic fractions
In groups, simplify the following;
66
67
a.
1
12
÷
1
4
b.
4x ÷
2
3
Present your answers to class. Discuss the example given below.
Example 4:
Expressing to lowest terms
Express to its lowest term
qp
yx
pq
yx
2
3
3
32
12
10
8
5
÷
Solution:
Invert the fraction on the right hand side and put
multiplication sign and then proceed in multiplicaion.
qp
yx
pq
yx
2
3
3
32
12
10
8
5
÷
=
yx
qp
pq
yx
3
2
3
32
10
12
8
5
×
=
xq
py
2
2
4
3
Now do the exercise below.
Exercise 4d
Simplify the following;
(a)
2
2
2
2
bc
a
bc
ab
÷
(b)
bd
a
cd
ab
7
4
5
6
2
÷
(c)
2
2
155
3
s
p
rs
pq
÷
(d)
ba
xy
ab
yx
2
2
4
10
8
5
÷
(e)
qs
p
rs
pq
7
4
5
6
2
÷
68
69
Simplication of algebraic fractions involving factorisation
Some simplication involves factorisation as the case is in (d)
Example 5:
Simplifying algebraic fractions
(a)
x
xx
3
2
2
+
Factorising the numerator, see that x is common and can be
cancelled out.
i.e.
x
xx
3
2
2
+
=
()
x
xx
3
2+
=
3
2+x
(b)
4
2
2
−
+
x
x
Here you factorise the denominator i.e. the difference of two
squares.
()()()()
2
1
22
2
22
2
4
2
2
−
=
−+
+
=
−+
+
=
−
+
xxx
x
xx
x
x
x
(c)
12
16
2
2
−+
−
xx
x
Here factorise both the numerator and the denominator and
cancel out common factor.
12
16
2
2
−+
−
xx
x
=
()()
()()
34
44
−+
−+
xx
xx
=
3
4
−
−
x
x
Exercise 4e
Simplify the following;
68
69
1. (a)
2
63
+
+
x
x
(b)
()
()()
32
3
−−
−
xx
xx
(c)
()
()()
55
5
+−
+
xx
xx
(d)
2
1
1)( +1)
(e)
()
()()
3232
323
−+
−
pp
pp
2. (a)
()()
33
3
2
−+
−
xx
xx
(b)
()
45
4
2
++
+
xx
xx
(c)
65
2
2
2
++
+
xx
xx
(d)
1
2
2
−
−
x
xx
(e)
12
4
2
2
−+
+
xx
xx
(f)
49
7
2
2
−
−
x
xx
Now you will look at some equations
Equations Involving Algebraic Equations
In this section you will learn how to solve linear equations in
our JCE mathematics and how to solve quadratic equations in
unit one of this book. In this section, you will learn how to solve
equations which involve algebraic fractions.
Activity 5:
Solving algebraic equation
Solve
3
10
32
=+
xx
Report your ndings to class. Now look at the examples given
below.
Example 6:
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71
Solving equations involving fractions
Solve the following equations
(a)
6
5
32
=+
xx
The rst step in solving such equations is to get rid of the
denominators by multiplying each term by the LCM of the
denominators. The resulting equation can then be solved by
the methods learnt previously.
Now the LCM of the denominator of 2, 3 and 6 is 12
∴
6
5
32
=+
xx
=
12
1046 =+ xx
You can ignore the LCM after this step
i.e. 6x + 4x = 10
10x = 10
x = 1
(b)
3
3
2
3 −
=
+ xx
The LCM of 2 and 3 is 6 and we have
()()
6
3233
3
3
2
3 −=+
=
−
=
+ xxxx
3x + 9 = 2x – 6 (ignoring the denominator)
3x − 2x = − 6 – 9
x = −15
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71
Exercise 4f
Solve the following equations
1. (a)
2
1
85
2
+=
xx
(b) 3x +
8
3
= 2 +
3
3x
(c)
8
5
23
1
4
3
+=+
xx
(d)
37
3
3
2
7
5 xx
−=−
(e)
5
3
3
1
2
1
=+
xx
2. (a)
3
3
2
3 −
=
+ xx
(b)
4
11
2
2
2
23
=
−
−
+ xx
(c)
2
4
23
5
2
=
−
−
+ xx
(d)
3
5
7
=
−m
(e)
2
3
3
29
4
53 −
=
−
−
− xxx
Real life problem of algebraic fractions
Sometimes, the equations are given in words and you are
required to write the equation and then solve it for the
unknown.
The following procedure will help you to construct linear
equations.
1) Represent the quantity to be found by a symbol (x is usually
used).
2) Make up the equation, which conforms to the given
information.
3) Make sure that both sides of the equation are in the same
units.
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73
Activity 6:
Solving algebraic fraction involving word problems
In groups, do the activity;
I think of a number, take
3
1
of it and then add 4. The result is 7.
Find the number I rst thought of.
Let one member from your group go and see the work of other
groups and see what they have done and report to the rest of the
members in the group.
Did you it in this way?
Let the number be x
Then
3
1
of x + 4 = 7
i.e.
3
x
+ 4 = 7
This is the equation and you can solve it using methods
learnt
i.e.
3
x
= 3
x = 3
×
3 = 9
x = 9
Did you get this? Now look at the example below.
Example 7:
Word problems involving algebraic fractions
Mary uses
12
1
of her salary to pay her bills, gives
15
1
to her
parents and she is left with K 6000.00. What is her salary?
Let her salary be x
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73
∴
6000
1512
=−−
xx
x
Then proceed as follows;
1
6000
15121
=−−
xxx
Note that the denominator of x and 6000 is 1
The LCM of 1, 12 and 15 is 60
i.e.
1
6000
15121
=−−
xxx
60
3600004560 =−− xxx
51x = 360000
x = 7058.82
Can you now practice formulating algebraic equations from the
exercise given below?
Exercise 4g
Construct equations from the following cases and solve them to
nd the value of the unknown.
(a) I think of a number, divide it by 4, add 5 and the result is 9.
What is the number that I rst thought of?
a) I think of a number, divide it by 3, subtract 4 and the result is
29. What is the number that I rst thought of?
b) I think of a number. When
7
3
is subtracted from
4
3
of the
c) number the result is 1. Find the number.
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75
d) When
2
1
is subtracted from
3
2
of a number the result is 4.
e) Find the number.
d) I think of a number. When
15
1
of this number is subtracted from
5
2
of the number the answer is
9
5
. What number did I
e) rst think of?
f) Find the number which, when added to the numerator and
denominator of the fraction
5
3
, gives a new fraction which is
equal to
5
4
.
Unit summary
In this unit, you have looked at algebraic fractions, simplifying
these by adding, subtracting, multiplying and dividing. The unit
has extended by expressing algebraic fractions to their lowest
term and simplifying algebraic fractions. In the next unit, you
will look at sets.
Unit review exercise
Evaluate and simplify if possible the following:
1 a.
1
+
1
2
b.
4
3
2
2
c.
2
+
3
d.
4
2
3
+
5
74
75
e.
4
3
5
7
f.
1
1
1)
2. a.
2
×
2
2
3
×
4
5
b.
×
2
3
×
4
5
c.
÷
3
2
d.
5
6
2
÷
3
5
e.
8 +12
12
2
27t
2
Glossary
An algebraic fraction is a fraction whose numerator and
denominator are algebraic expressions.
g.
1
2
5 +6
+
3
h.
1
i.
+1
(
1)
–
1
+1
j.
2 +5
3
+
3 +2
6
k.
2+
3
5
6
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77
References
Chikwakwa, e t al (2002), Senior Secondary Mathematics
Student′s Book 3. Blantyre: Macmillan Malawi.
Hau S and Saiti F (2010). Strides in Mathematics 3. Blantyre:
Longman Malawi.
G D Buckwell and B N Githua, KCSE Golden Medal
mathematics, Macmillan, London.
Gunsaru, et al, Secondary Mathematics Book 3, Dzuka
Publishing Company, Blantyre Malawi.
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77
Unit
5
SETS
In form 1, you learnt that a set
is a collection of objects. In this
unit, you will learn more about
sets . By the end of this unit you
must be able to describe elements
of a set and representsets in Venn
diagrams. The idea of sets is used
in many elds like manufacturing
and other elds.
In everyday life we buy things in
sets.
Describing elements of a set
A set is a collection of objects. The
things or objects that make up a set
are called members or elements
of the set. It is possible to have a set
which contains objects that do not
have anything in common. What
does mathematical instruments box
contain? You may nd objects such as
a compass, divider, ruler, protractor
and set square. Whether the objects
are related in some way or not, as
long as they form a collection, they
comprise a set. Now do the activity
below.
Activity 1:
Describing set builder notation
In pairs
1.Identify objects or things in a set in
your class
2.Write set of objects in set language.
3.Write down the symbol for the
following:
Universal set
Not a member of
A union B
A intersection B
Share your work with your friends
and discuss your ndings. Now look
these in detail.
There are various symbols that are
used in set language. Now look at the
followings is a member of, ∉; is not
a member of or { }; the empty set or
null set;
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79
B ⊂ C; B is a proper set of C
A ⊆ B; A is a subset of B
A ⊇ B; A contains B
⊄9; the negation of ⊂, ⊆ and ⊇
A ∪ B; union of A and B
A ∩ B; intersection of A and B
n(A); number elements in set A
A΄; complement of set A
Set builder notation is another way of representing sets.
Inside the set, unknown or variable is used to represent the
elements of a set in general. This is followed by a vertical line
or a colon which means “such that”. The vertical line or colon is
followed by a description of the elements. Look at the examples
below.
Example 1:
Set builder notation
Write the elements of the set given in set builder notation.
a. T = {t│t is all teachers at your school} or T = {t: t is all
teachers at your school}.
A = {Chimtengo, Chitedze, Chinthuzi, Chimwemwe,
Getrude}
b. B = {y : y < 10, y is a whole number} or B = {y │ y < 10, y is
a whole number}
B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
c. X = {z │−5< z < 4, z is an integer} or X = {z : −5< z < 4, z is
an integer}
X = {−4, −3, −2, −1, 0, 2, 3}.
Hence the elements in set X are integers greater than −5
and less than 4. Sets A, B, C are all nite sets. Now do the
following exercise.
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79
Now do the exercise below.
Exercise 5a
1. If A = {integers}, list the members of the following sets
using… where appropriate.
(a) {x : x 9}
(b) {y │ y}
(c) {a : a ≥ −3}
(d) {x : x 0}
(e) {y : y – 5 = 0}
(f) {x : 2x + 4 = 16}
(g) {y : − 7 ≤ y 5}
(h) {x : − 8 x − 1}
(i) {x : x 0 and has no remainder}
2. List the elements of the following sets:
a. D = {d: d is public holiday days in Malawi}
b. E = {e│e is even numbers less than 20}
c. G = {g : g ≤ 18, g is an odd number}
d. Y = {x : x is counting numbers less than 10}
e. H = {d │d is district along the lake Malawi}
3. If D = {1; 2; 3; 4; 5; …20}, list the members of the following
sets.
a. {a : a is a square number, a
D
b. {b│ b +2 15 , b D}
c. {x : x is a factor of 36, x
D}
4. Give that {a, c, e, h, I, l, m, s, t, w} is a universal set and
Y = {a, c, e, h, i, m, s}, nd n(Y’).
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81
Denitions
Equal sets are sets that contain exactly the same members,
regardless of the order in which the members are presented.
If A = {a, b, c, d, e} and B = {b, c, e, d, a} then A and B are equal
sets.
Equivalent sets have the same number of elements.
If H = {1, 2, 3, 4, 5} and G = {a, d, e, f, g}, how many elements has
G and H? These two sets are equivalent.
A subset contains wholly or part of the elements in a universal
set.
If M = {a: a is whole number less than 10} and N = {1, 2, 3, 4, 5}
then N is a subset of set M.
A set is said to be proper subset if it has few element than
those given in the universal set or if it is not exact subset of
itself. From the above statement N is a proper subset of M
An improper subset is a subset which has all the elements in
the original set. It is equal to the original set.
A set is said to be nite if the elements in there can be counted
for example;
M = {a, c, b, g}, M has four elements while innite set the
elements in that particular set cannot be counted for example;
N = {1, 2, 3, 4, 5, 6…}; a set of natural numbers,
P = {1, 3, 5, 7, 11, 13, 17,.}; a set of prime numbers,
Z = {−4, −3, −2, −1, 0, 1, 2, 3…}; a set of integers.
The dots show that the elements continue.
The number of elements in a set is called the cardinal number
of the set.
Empty or null set is a set that has no members. For example;
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81
let B be all boys in class, G all girls in class and H neither a girl
nor a boy. This can be also shown in this way;
B
G
H is an empty set since you can hardly nd a person who has
both sex. B and G are disjoint sets.
Oral exercise
Given the following sets; A = {2, 4, 5}, B = {1, 2}, C = {5, 2, 4},
D = {a, c,} and E = {x: x is a number greater than 1}
Identify sets that are equal, equivalent, nite and innite.
From exercise 5a above, you are able to come up with subsets
from the main set. What do you call the main set? You are going
to look at that now.
Universal set
The universal set is the set which contains all the possible
elements. If
A = {y: y is a letters the alphabet} and B is a set of vowels of the
alphabet then A = {a, e, i, o, u}, therefore A is the universal set
and B is a subset of A.
A teacher, boys and girls in a class form a universal set of people
present in a classroom. In this girls only or, boys only could be a
subset of this set. Universal set is denoted ξ. Diagrammatically
sets A and B could be represented as follows:
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83
A
B
A= {a, b, c… z} and B = {a, e, i, o, u}
B is a proper subset of A or B ⊂ A.
Can you now come up with your own universal set?
Complement of a set
The complement of a set is the set containing all elements in a
universal set but are not members of this given set.
In other words, the complement of B is the set which contains
all those elements that are in a universal set,
U
but do not
belong to set B. The complement of a set B is denoted B.
B
The shaded region represents the complements of set B denoted
B´
Example 2:
Complement of a set
a. = {1,2,3,4,5,6,7,8} and A = {2,4,6,8}. What set is
represented by A?
Solution:
U
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83
A consists of all elements in
which are not in set A.
∴
A′ = {1, 3, 5, 7}
Show this diagrammatically as demonstrated above.
b. If
= {all triangles} and Q = {equal sided triangles}. What
set is represented by Q′?
Solution:
= {Scalene, Equilateral, Isosceles
s}
Q = {Equilateral
s}
∴
Q′ = {Scalene, Isosceles}
Note: A universal set,
is the background set i.e. it contains all
the possible elements. If a certain set Q, for example, contains
some elements of the universal set, then Q is said to be a subset
of . But when it comes to Q, this means a set containing
elements of the universal set which are not however contained
in set Q.
Now do the exercise given below.
Exercise 5b
1. Let
U
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 2, 5, 7, 9}.
What set is represented by A′?
2. Let
U
= {all quadrilaterals} and B = {quadrilaterals with
all 4 sides equal}. What set is represented by B′? Give 2
examples of elements of B and give 3 examples of elements
of B.
3. If
U
= {m, a, t, h, s} and C = {a, h, s}, what set is
represented by C’?
4. If
U
= {1; 2; 3; 4; …10}, A = {1; 2; 5; 7}, B = {1; 3; 6; 7}, write
down the sets A’ and B’.
5. Given that the universal set
U
= {1, 2, 3, 4, 5, 6, 8, 9},
sets N and M of
U
such that N = {even numbers} and M =
{perfect square}, nd;
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85
(a) N΄
(b) M΄
(c) n(N΄)
Union of sets
The union of two or three sets is everything which belongs to
either, both or all the three sets and represented by the symbol
.
Activity 2:
Identifying elements of a union of two or three sets
In pairs,
(a) Get three mathematical boxes or sets of playing cards.
(b) Label them as set A, B and C.
(c) List the elements in each set.
(d) Write all elements in the three sets as a new set.
(e) Express the new set in terms of A, B and C.
(f) List A.
Present your work to the class. Look at the case below.
Consider A to be a set of all vowels of the alphabet. Let A = {a, e,
i, o u} and B to be a set of whole numbers from 1 to 5. Then
B = {1, 2, 3, 4, 5}.
Now write all the elements that are in A and B. You obtain the
following set {a, e, i, o u, 1, 2, 3, 4, 5}. This is a set of all elements
of A and B. Therefore, this is known as Union of Sets A and B.
In short A ∪B = {a, e, i, o u, 1, 2, 3, 4, 5}.
Example 3:
Union of sets
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85
a. Let A = {a, b, c, d}, B = {a, 1, d}. What is AB?
Solution:
A ∪ B = {a, 1, b, c, d}
Note: a and d are not written twice since they are same
elements.
b. Let C = {Malawi, Zambia, Zimbabwe}
D = {Zomba, Thomas, Hanna}. What is CD?
Solution:
C ∪ D = {Malawi, Zambia, Zimbabwe, Zomba, Thomas,
Hanna}
Exercise 5c
1. Let = {a, b, c, d, e, f}, X = {a, d, e}, Y = {a, b, e} and Z = {a,
b, f}. List the elements of the following sets.
(a) X ∪Y
(b) X ∪ Y ∪ Z
(c) Y ∪ Z
(d) X ∪ Z
(e) X΄ ∪ Z΄
2. If
= {a; b; c; d; e}, A = {a; c; e} and B = {a; e}, list the
members of the following sets.
(a) A ∪ B
(b) A΄ ∪ B
(c) A΄ ∪ B΄
(d) n(A ∪ B)
(e) (A ∪ B)΄
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87
3. If = {1, 2, 3, 4, 5}, N = {1, 3} and M = {3, 4}. Find
(a) N΄
(b) M΄
(c) (N ∪M)
4. Given that the universal set
U
= {1, 2, 3, 4, 5, 8, 9}, sets A
and B of
U
such that A = {even numbers} and B = {perfect
square}, nd;
(a) A
(b) B
(c) n(A ∪ B΄)
(d) A΄ ∪ B΄
(e) n(A΄ ∪B΄)
Having looked at union of sets you will now look at intersection
of sets.
Intersection of sets
Intersecting sets are those sets which have some common
elements in them. These sets can be two or more.
Activity 3:
Identifying elements of in an intersection of two or three
sets
In pairs,
(a) Get three mathematical boxes or a set of cards of numbers.
(b) Label them as set A, B and C
(c) List the elements in each set
(d) Write elements common in the three sets, A, B and C.
(e) Express the new set in terms of A, B and C.
U
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87
Present your work to the class. Look at the case below.
Consider set C = {rst four letters of alphabet} and set D = {c, m,
x, d}.
These sets are not equal.
The symbol
∈
means “belongs to” or “a member of” e.g. Zomba
∈
D, means Zomba belongs to D.
Set C = {a, b, c, d,}.
You see that a
∈
C, b
∈
C, c
∈
C and d
∈
C, also c
∈
D, m
∈
D, x
∈
D
and d
∈
D.
Thus there are elements which are common to both sets. As
already stated these are intersecting sets and symbol for
intersection of sets is ∩.
Therefore C ∩ D = {c, d}. These are the only two elements, which
are found in both sets C and D.
Example 4:
Intersection of sets
Let A = {a, b, c, d, 2, 3}
B = {1, 2, 3}
C = {p, q, r, t}
Find (a) A ∩ B
(b) A ∩ C
Solution:
(a) A ∩ B = {2, 3}
Since 3
∈
A, 3
∈
B, 2
∈
A and 2
∈
B.
(b) A ∩ C = Ø, (Ø is the symbol for an empty set. In other
words, there is nothing in the given set).
Since A and C do not have any common element
Let P = {all positive whole numbers from 5 to 10}
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89
Q = {even numbers between 2 and 16}
R = {natural odd numbers}
Find (a) P ∩ Q
(b) Q ∩ R
(c) P ∩ R
Solution:
(a) P = {5, 6, 7, 8, 9, 10}
Q = {4, 6, 8, 10, 12, 14}
R = {1, 3, 5, 7, 9, 11, 13, 15,…}
∴
P ∩ Q = {6, 8, 10}
(b) P ∩ R = {5, 7, 9}
(c) Q ∩ R = Ø since there are no even numbers which are odd
numbers and there are no odd numbers which are even.
Exercise 5d
1. Let ξ = {a, b, c, d, e, f}, X = {a, d, e}, Y = {a, b, e} and Z = {a, b,
f}. List the elements of the following sets.
(a) X ∪ (Y ∩ Z)
(b) X ∩ Y ∩ Z
(c) (X ∪ Z) ∩ Y
(d) X ∩ Z΄
2. Let P = {2, 3, 5, 7, 11, 13, 17}. List the elements of the
following sets:
(a) P ∪ Ø
(b) P ∩ Ø
3. If ξ = {a; b; c; d; e} and A = {a; c; e} and B = {b; e}, list the
members of the following sets.
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89
(a) A΄ ∩ B
(b) B΄ ∩ B
(c) (A ∩ B)΄
(d) A΄ ∩ B΄
(e) (A ∪ B)΄
(f) A΄ ∪ A
4. If ξ = {days of the week}, S = { words which contain the
letters s} and N = {words which contain six letters}.
(a) List the members of the sets S, N, S΄, N΄;
(b) List the members of the set,
(i) (S ∪N)΄
(ii) (S ∩ N)΄
(c) Hence, without further working, list the members of
the sets
(i) S΄ ∩ N΄
(ii) S΄ ∪N΄
5. If ξ = {c; h; I; d, k; e; n}, P = {n; i; c; e}and Q = {h; e; n}, list
the elements of the following;
(a) P ∩ Q
(b) P ∪Q
(c) (P ∪ Q)΄
(d) (P ∩ Q)΄
(e) P΄ ∩ Q
(f) P ∪ Q΄
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91
Number of elements in a set
In pairs discuss
If set A = {1, 3, 4, 5, 6, 7, 9} and set B = {2, 3, 4, 6}. List the
following;
(a) A ∪ B
(b) A ∩ B
(c) n (A)
(d) n (B)
(e) n (A ∪B)
(f) n (A ∩B)
Show that n (A) + n (B) = n (A ∪B) + n (A ∩ B).
From this activity, you might have seen the following;
n (A) = 7; number of elements in set A
n (B) = 4; number of elements in set B
n (A ∪ B) = 8; total number of elements in A and B
n (A ∩ B) = 3; number of elements common in both A and B.
Substituting; n (A) + n (B) = n (A ∪ B) + n (A ∪ B)
This gives 7 + 4 = 8 + 3 = 11
Therefore for given sets A and B; n(A) + n(B) = n(A ∪B) +
n(A ∩B) usually written as n(A ∪B) = n(A) + n(B) − n(A ∩B)
Venn diagrams
Venn diagrams are the principal way of showing sets
diagrammatically. The method involves mainly entering the
elements of a set into a circle or circles. You can use Venn
diagrams for both union and intersection of sets.
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Activity 4:
Illustrating elements of sets in a Venn diagram.
In pairs discuss; If B = {0, 2, 3, 5, 7, 10}, C = {2, 3, 4, 5, 6, 8, 9}
and
D = {0, 1, 2, 8, 10, 13, 15}. Write the elements in the appropriate
set in the Venn diagram below;
B
C
D
Write the set language for the listed elements above as follows;
(a) Elements found in B and C only
(b) Elements common in B, C and D
(c) Elements found B and D only
(d) Elements found in B only
(e) Elements found in C only
(f) Elements found in D only
Present your work to the whole class. Now you will look at this
in detail.
Intersection of Sets
Consider the sets, A = {a, b, c, d, e} and B = {1, 2, 3, d, e,x}.
92
93
A
a
e
c
d
b
All the elements of set A are inside the circle. Now look at the
following sets:
A = {a, b, c, d, e} and B = {1, 2, 3, d, e, x}.
These sets can be represented as follows:
A
B
1
2
3
x
a
b
c
d
e
In this diagram, it can be seen that those elements which belong
to both sets are placed in the region of the two circles. This
overlapping shows that the two sets intersect i.e.
A ∩ B = {d, e}.
What happens to two sets A and B whose intersection is empty
i.e.
A ∩ B = Ø? The Venn diagram looks like this:
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93
A
B
The two circles do not intersect. This shows that the two circles
have no element which is common to both of them.
Example 5:
Venn diagrams
In a form 3 class of 108 students, 60 students like football, 53
like volleyball and 10 like neither. Calculate the number of
students who like football but not volleyball.
Solution:
Let the Venn diagram be like this:
10
F
60 – x
53 – x
x
W
Let the number of students who like both volleyball and football
be x.
∴
Students liking football only is 60 − x and those liking
volleyball only is 53 − x.
See the diagram above
∴
60 − x + x + 53 − x = 98
−x = 98 −60 −53
94
95
−x = 98 −113
∴
x = 15 divide both sides by −1
∴
Students likes football but not volleyball = 53 − 15 = 38
Exercise 5e
1. If A = {prime numbers less than 20} and B = {1, 2, 3, 4,..10}
draw a Venn diagram illustrate the relation between A and
B.
2. If M = {all integers from 2 to 15} and N = {Prime numbers
less than 20}
(a) Draw a Venn diagram to illustrate the information
above
(b) List the elements of the set M
N.
3. P = {1, 4, 7, 11, 15, 17}
Q = {5, 10, 15}
R = {1, 4, 9}
Represent this information on a Venn diagram.
4. Let A = {months of a year}, B = {months starting with M}
and C = {months starting with N}
(a) List elements of A, B and C
(b) Draw a Venn diagram to illustrate
(i) A∩ B
(ii) A ∩C
(iii) B ∩ C
5. Given that the universal set ξ = {1, 2, 3, 4, 5, 8, 9}, sets A
and B of ξ such that A = {even numbers} and B = {perfect
square}, nd;
Find the following;
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95
(a) n(A ∩B)΄
(b) Show that n(A ∪B) = n(A) + n(B) − n(A ∩ B)
6. The number of elements in each region of the Venn diagram
in the gure below are given as shown.
M
N
P
6
2
4
y
11
5
If n (ξ) = 36, nd
(a) y
(b) n(N)
(c) n(M ∩ P)
(d) n(N΄ ∩ P΄)
7. Given that P and Q are sets such that n(P) = 9 and
n(Q) = 13 and n(P ∩ Q) = y, nd n(P΄ ∪Q΄) in terms of y. If
the total number of elements in P and Q is 15, nd the value
of y.
8. In the Venn diagram below, the numbers of elements are
shown.
A
C
4 + a
15
9
a
B
14
If the n( ) = 9a, nd
(a) a (b) n(A) (c) n(A ∪ B)
(d) n(A ∩ B΄)
U
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97
Union of Sets
Union of sets can also be presented in Venn diagram. Firstly do
the activity below. Venn diagram is also used in Union of sets.
Activity 5:
Illustrating union of sets in the Venn diagram
In pair, discuss;
If A = {d, e} and B = {a, b, c, d, e}, illustrate the given set in the
Venn diagram;
Report your work to class.
Do more solving problems involving Venn diagram
Solving problems involving sets using Venn diagram
Venn diagrams in mathematics are also used to solve problems
which might be more difcult to solve without the use of Venn
diagrams. See the example given below.
Example 6:
Using Venn diagrams
In a group of 20 college students, 12 are taking Mathematics, 10
Physics and 14 Chemistry. 6 take Mathematics and Physics, 4
Chemistry and Physics and 8 Mathematics and Chemistry. Each
student is taking at least one of these subjects. How many of the
students are taking all the three subjects?
Solution
Let M = {students taking mathematics},
P = {students taking Physics}
C= {students taking Chemistry}
Illustrating as a Venn diagram;
The universal is ξ = M ∪ P ∪ C.
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97
M
P
C
y - 2
6 - y
y
8− y
4 - y
y + 2
y
n (M ∩ P) = 6– y; students taking Mathematics and Physics
only.
n (P ∩C) = 4 − y; students taking Physics and Chemistry.
n(C ∩M) = 8−y; students taking Mathematics and Chemistry.
n(M ∩P ∩C) = y; students taking all the three subjects.
n (M) = 12; students taking Mathematics, n (M ∩P΄ ∩C΄);
Students taking Mathematics only
= 12 – (6 – y) – y – (8 – y)
= 12 – 6 + y – y – 8 + y
= y – 2
n (P) = 10; students taking Physics
n(P) = P ∩M΄ ∩C΄ = Students taking Physics only
= 10 – (6 – y) – y – (4 – y)
= 10 – 6 + y –y – 4 + y
= y
n( C ) = 14; students taking Chemistry
n(C ∩P΄∩M΄); Students taking Chemistry only
= 14 – (8 – y) – y – (4 – y)
= 14 – 8 + y – y – 4 + y
= y + 2
98
99
But n(M ∪ P ∪ C) = 20; the total number of students
(y – 2) + (6 – y) + y + (8 – y) + y + (4 – y) + y + 2 = 20
9 +y = 20
y = 20 −18
y = 2
n (M ∩P ∩C) = 2; students taking all the subjects.
Now do the following exercise.
Exercise 5f
1. Use the Venn diagram to answer the questions below.
2
8
4
6
A
B
5
1
List the elements of the following
(a) ξ
(b) A
(c) A ∩B
(d) A ∪B
(e) (A ∪B)΄
(f) (A ∩ B)΄
2. In a group of 100 students, 55 likes netball, 63 likes rugby
and 15 likes neither, calculate the number of students who
like only netball.
3. In a class of 30 students 20 takes History, 15 French and 2
(a) ξ
98
99
takes neither. How many take;
(a) both History and French?
(b) History only?
(c) French only?
4. Let B = {all boys in your class} and G = {all girls in your
class}. Draw a Venn diagram to illustrate B ∩G.
5. Consider the gure below:
A
B
11
18
14
2
16
20
3
5
1
10
7
List the elements of
(a) ξ
(b) A
(c) B (d) A ∩B
(e) (A ∩B)΄ (f) A΄
(g) B΄ (h) A ∪B
6. The gure below shows a Venn diagram.
N
V
2x
2x+1
x
10
In the Venn diagram, ξ = {girls in form three}
N = {girls that play netball}
V = {girls that play volleyball}
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101
Given that there are 21 girls in the class, nd how many
girls play both netball and volleyball.
7. Given that universal set ξ = {11,14,15,17, 18, 20, 23, 26},
and set X = {11, 14, 15, 17, 18, 20} and set Y = {15, 17, 18,
20, 23, 26}, nd X΄ ∪ Y΄
8. Given that the universal set ξ = {11, 12, 13, 14, 15, 16, 17,
18, 19}, Set B = {number greater than 16}, and set
C = {multiples of 3}.
Find the elements of:
(a) Set B
(b) Set C
(c) Set (B ∪C)΄
Unit summary
• A set is a collection of objects. The things or objects that
make up a set are called members or elements of the set. It
is possible to have a set which contains objects that do not
have anything in common. Whether the objects are related
in some way or not, as long as they form a collection, they
comprise a set.
• Sets are denoted by a capital letters. There are various
symbols that are used in set language. In the next unit, you
will look at mapping and functions.
Unit review exercise
1. A class of 50 students wrote tests in mathematics, biology
and physical science. The results of the tests were as
follows; 12 passed mathematics and physical science, 19
passed mathematics and biology; 17; passed Biology and
physical science; 2 passed physical science only, 5 passed
mathematics only and 6passed Biology only. If 5 students
failed all the three subjects and y passed all the subjects,
use Venn diagram to calculate the value of y.
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101
2. Given that A = {11, 12, 15,17}, B = {11, 15, 17, 19, 20} and
C = {11, 17, 21}.
(a) Show the three sets on a Venn diagram
(b) Find n(A).
3. Given that X = {a, e}; Y = {b, c, d, e} and Z = {c, d, e, f},
nd {X ∪ Y ∪ Z}
4. The gure below is a Venn diagram showing the number of
elements in sets M, N and universal set ξ.
M
N
4x
4−x
16
2x
If n(M ∪ N) = 29, calculate the value of x.
5. At Chimwemwe secondary school, 72 students like watching
football, 64 like watching basketball and 62 like watching
netball, 18 like watching football and basketball, 24
like watching football and netball and 20 like watching
basketball and netball, 8 students like watching all the
three games and 56 don’t like watching any game.
(i) Draw a Venn diagram representing this information.
(ii) From the Venn diagram, calculate the number of
students at the school.
6. A church congregation has a youth group and a music
group. There are 400 people in the congregation out of
which 40 people belong to both the youth group and music
group. There 60 members who belong to the youth group
only, while 220 belong to neither the youth group nor the
music group.
(i) Draw a Venn diagram illustrating this information.
(ii) Calculate the number of people who belong to the
music group only.
7. In a family, six members eat meat; ve members eat sh
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103
while two members eat both. Calculate the number of
members in the family.
8. In form 3 class, students learn French, Latin and History.
20 students learn French, 55 learn Latin and 37 learn
History. 7 students learn French and Latin only, 5 learn
Latin and History only, 2 learn French and History only, 10
do not learn any of these subjects while x students learn all
the three subjects. If there are 100 students in the class,
(a) Draw a Venn diagram to represent this information.
(b) Use your Venn diagram to calculate the number of
students who learn Latin only.
9. The gure given below shows a Venn diagram of sets A, B
and C.
5
3
12
2
7
4
9
A
B
C
Find A’ (B C).
(i) Draw a venn diagram and shade the region
representing A΄ ∩ B΄ ∩C.
(ii) Find n(A΄ ∩B΄ ∩ C), if n(A ∪ B) = 8 and
n(A ∪ B ∪C) = 12
10. In a class of 50 students, each of the students ate at least
one of the following types of fruits: Banana, mango, and
orange. It was found that: (x + 1) students ate all the three
types of fruits, 9 students ate mangoes and oranges only, 8
ate bananas and mangoes only, 5 ate banana and oranges
only, x students ate bananas only, (x − 1) students ate
mangoes only and (x + 4) students ate oranges only.
(i) Illustrate the information using a venn diagram.
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103
(ii) Find the number of students who ate mangoes.
11. During National Examinations, 37 students sat for an
examination in Mathematics, 48 Physical science and 45 sat
for Biology. 15 students sat for Mathematics and Physical
science. 13 sat for Physical science and Biology, 7 sat for
Mathematics and Biology and 5 students sat for all three.
Draw Venn diagram to represent this information and
hence nd the total number of candidates.
Glossary
A set is a collection of objects.
Empty or null set is set that has no elements.
Universal set is the main set that contains all the elements in
it.
A Subset is a set that contains wholly or part of the elements in
a universal set.
Equal sets are sets that contain exactly the same members,
regardless of the order in which the members are presented.
A proper subset is a set that has few elements than those
given in the universal set or if it is not exact subset of itself.
Finite set is a set with countable number of elements in it.
Innite set is set whose elements are not countable
Union of set is the set of all elements that are of either.
Intersection of set is set that has elements that are common
in all the given sets.
The complement of a set is the set containing all elements in
a universal set but are not members of this given set.
References
Chikwakwa, e t al (2002), Senior Secondary Mathematics
Student′s Book 3. Blantyre: Macmillan Malawi.
104
105
Hau S and Saiti F (2010). Strides in Mathematics 3. Blantyre:
Longman Malawi.
J B Channon et al, New general Mathematics 3, A modern
Course for Zimbabwe, Longman group Ltd (1996), UK.
Elaine Ryder et al, CHANCO Teach yourself series, Mathematics
Questions and Model answers, second Edition (2013), Chancellor
College Publications, Zomba, Malawi
104
105
Unit
6
MAPPING AND FUNCTIONS
In the last unit you looked sets.
In this unit you are going to look
at functions. A function is a value
which depends on and varies with
another value. You are going to
represent functions in different
forms and also nd the range and
domain of functions.
Functions are the central object
of investigation in most elds of
modern mathematics.
The ideas of functions are used
on daily basis knowingly or
unknowingly. Functions will help
to promote your thinking.
Dening mapping and
functions
Before dening mapping and
function, you will rst of all do the
activity.
Activity 1:
Identifying relations
a) Make two groups ,A and B of
seven students each.
b) Each student in A should carry
a card with a name of one
student in group B.
c) Using a rope match a name in
B with every person in A.
(i) To how many persons is
the rope being assigned?
(ii) What do you call such
linking?
From the above activity, you might
have seen that every person of group
A has a unique corresponding name
in set B. Such a correspondence is
called a relation.
Such kind of a correspondence also
happens with numbers.
For example, if set A is all real
numbers and set B is twice the
number in A then this denes a
relation between elements set of A
and those of B. In this case;
1 corresponds to 2,
3 corresponds to 6,
4 corresponds to 8,
6 corresponds to 12
106
107
Notice that every number in set A corresponds to a unique (only
one) number in set B in other words; each element in A maps
onto one element in B.
Can you list the ordered pairs from the example above and come
with own ordered pairs.
Thus a relation is a set of ordered pairs such as (1, 2), (3, 6),
(4, 8) and (6, 12) from above.
The set of ordered pair has input and output values. The set
of inputs is called domain and the set of output is called the
range also referred to as image.
Relation
There are of four types relations namely;
1. One - to - one relation
One member in a domain relates or maps onto one member
of the range. e.g. Square the input to get output.
A B
1 1
2 4
3 9
4 16
2. One – to – many relation
A member of the domain A relates to more than one member
of range B. Students who belong to more than one club at
school.
John
Mary
Maths
Debate
CAPSO
SCOM
A B
3. Many – to – one relation
For one element in the range, there is more than one
106
107
element of the domain that relate to it. This is also called
multi-valued functions. For example, a number of students
may obtain the same grade in a particular subject in an
examination.
Debora
James 1
Harry
Meda 2
Rabecca
4. Many – to – many relation
For one element of the domain, there is more than one
member of the range that relates to it and one element in
the range, there is more than one element of the domain
relating to it. Students are members of more than one club
society and a club society a club society has more than one
student.
Dziko Science
Kafa Debate
Dala YCS
Example 1:
Identifying relations
Which of these arrow diagrams are relations?
(a)
A B
a d
be
c f
(b)
A B
a d
be
c f
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109
(c)
A B
0
1 2
2 3
8
In diagrams (a), (c) represent functions, why? This is because
every element in A has a unique element in B. Diagram (b) does
not represent a relation. Why?
In these diagrams, the elements in set A are called the inputs
and in set B are called the outputs.The set of inputs is called
domain and the set of output is called the range also referred
to as image
Exercise 6a
In the following diagrams, say whether the arrow diagrams
represent relation and if not explain why not?
(a)
a p
b q
c r
d s
(b)
1 3
2 5
3 9
usipa
(c)
2 3
3 7
9 17
5 24
(d)
utaka
chambo
mcheni
mcheni
sh
108
109
(e)
2 3
3
1
2
9
1
2
5
1
4
(f)
1 3
2
3
9
Mapping
A mapping is a special relation in which each element in one
set is related to one element of the other set or pairing of input
values with output values. Which of the four types of relation is
a mapping? Certain statements are used in mapping such as,
If y is the operation “the square of”, written as A
Y
B. In this
statement, elements in A are squared to get elements in set B.
Now look at the example below.
Example 2:
Elements of a relation
Let a be the operation of “add 3 to” A = {1, 3, 5, 7, 9}. What are
the members of the new set?
Solution
Let the image which is the new set be B
B = {4, 3, 8, 10, 12}; add 3 to input to get an output.
Exercise 6b
1. What type of relation is presented by the diagrams?
110
111
(a)
1 3
2 5
3 9
(b)
a 6
b 4
c 2
d
(c )
Chitipa
Nsanje
Mchinji
Mangochi
Malawi
2. B is the operation “multiply by 2, add 1”, G = {0, 1, 2, 3, 5}
and G
B
H. list the members of H.
3. X is the operation “square and add 2” A = {2, 3, 4, 5}. What
are the members onto which A is mapped?
4. V is the operation “cubed” and B = {1, 2, 3, 4, 5}. What are
the members of S, the set onto which the V is mapped?
5. Let S be the operation “the squaring of ˮ and T = {1, 2, 3, 4,
5, 6}. What is the image of T’?
Identifying functions
In mathematics a function is a relation between a set of
permissible outputs with property that each input is related to
exactly one output or
a rule which maps a single number to another single number.
An example is the function that relates each real number x to its
square x
2
. Can you generate this function?
Which of the four types of relations are functions?
Only one – to – one mapping and many – to − one mapping
represent function.
110
111
Oral exercise
Which of the following are functions? Give reason for your
answer.
domain range
domain range
(a)
3 9
5 25
7 49
(b)
a p
b q
c r
d s
domain range
domain range
(c)
Chitipa
Nsanje
Mchinji
Salima
Malawi
(d)
2 1
3
1
2
9
1
3
5
1
4
If a mapping is a function, it must satisfy these conditions;
(a) Each element of the domain has a unique image.
(b) The domain and the range have one – to – one
correspondence or have many – to – one correspondence.
Function notation
There are several ways of describing a function. They are by
means of;
(a) A mapping diagram like what you have done so far.
(b) an algebraic equation such as y = 5x−4
(c) A graph. y = x
2
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113
x
y
(d) a set of ordered pairs like {(0,1), (1, 2), (2, 3)}
(e) as a table of values
x 0 1 2 3 5
y 0 3 6 9 15
The algebraic functions notation can take three different forms
and these are;
(a)
45 −= xy
(b)
45)( −= xxf
read as the function f of
.45 −= xx
(c)
45: −→ xxf
.
which is read as the function f such that x
maps onto 5x – 4.
All the three expressions mean that y is a function of
)(),( xhxg
are also used to double functions.
A function f takes an input x, and returns an output f(x). x is the
value taken by the function when you evaluate f(x) at a point.
f(x) is commonly used.
One metaphor describes the function as a "machine" or as
"black box" that for each input returns a corresponding output.
Each element of the domain is said to be mapped onto the
element of the second set (output) that corresponds to it.
You put the value of x into machine to get the output value. See
the illustration.
112
113
Output, f(x)
Input x
f
Input and Output machine
Activity 2:
Finding the range given the domain
In pair, nd the range given that f(x) = 5x – 4 if domain is
{0, 1,2 ,3}
Present your work to class.
Now look at the example below.
Worked Examples 3:
Range and domain
a. Consider the function
53)( += xxf
with domain {5, 8, 11, 14,
17}. The range is found by substituting each number of the
domain into
.53 +x
Thus when
4
3
2
1
0
=
=
=
=
=
x
x
x
x
x
()
()
()
()
()
175)4(34
145)3(33
115)2(32
85)1(31
55)0(30
=+=
=+=
=+=
=+=
=+=
f
f
f
f
f
Hence the range is the set
5, 8, 11, 14, 17
In this example, 0 is mapped onto 5, then 5 is said to be the
image of 2, and 0 is a pre – image of 5
a) For the function f: x → x
2
−3
Evaluate:
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115
(i) f (1)
(ii) f (5)
(iii) f (–4)
(iv) f (0)
Solution
Substitute x by the given x- value.
f (1) = (1)
2
–3 = −2
f (5) = (5)
2
–3 = 22
f (−4) = (−4)
2
–3 = 13
f (0) = 6
2
−3 = −3
Exercise 6c
1. If
,32)1( += xf
evaluate:
a)
()
1f
b)
()
3f
c)
()
7f
d)
()
10f
e)
()
40f
2. If
()
,34 −= xxf
evaluate:
a)
()
0f
b)
()
2f
c)
()
5f
d) f(−1)
e)
f(a)
3. If
()
=xg
x
2
+ 6, calculate:
a)
()
2g
b)
()
4g
c)
g(−3)
d)
()
1g
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115
e) g(−1)
4. If
()
xxg 2=
2
−5
calculate:
a)
()
0g
b)
()
1g
c)
()
2/1g
d)
()
4/1
−g
e)
g(−3/2)
5. If
()
2
,
43
−
=
x
xh
calculate:
a)
()
0h
b)
()
4h
c)
()
6h
d)
()
10f
e) h(−2/3)
6.
()
4
23
+
=
x
xf
, calculate:
a)
()
6
−f
b)
()
5
.
2
f
c)
()
5
.
0
−f
d)
()
0f
e)
()
6
.
1
−f
7. If
4
86
:
+−
→
x
xh
, calculate:
a)
()
ih
b)
()
0
h
c)
()
4
h
d)
()
5
.
1
h
e)
()
22
−h
8. If
()()
x
xx
x
−
−+
→
42
, calculate
a)
()
1g
b)
()
4g
c)
()
0g
d)
()
2
−g
e)
()
8
−g
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117
9. For the function
3
2
:
−−
→
x
xf
. Evaluate
a)
()
4f
b)
()
10f
c)
()
8
−f
d)
()
0f
e)
()
1f
10. If
()
xxf
3
=
2
8
5
−− x
, write down the value of
a)
()
2
−f
b)
()
1f
c)
()
0f
d)
()
2f
e)
()
4f
11. Copy and complete gure 6.1 if the relation is
a)
23
−→ xx
b)
xx →
2
1
2
3
4
5
1
Figure 6.1
12. With {1, 3, 4, 5, 6} as domain draw mapping diagrams for
23
:
+→ xxf
and
xxf
3
:
→
2
2
−
Finding the domain given the range
Sometimes you know the range and you are asked to nd the
domain. To do this you solve the equation for x.
Activity 3:
Finding the element in the domain
In pairs do the following, If f(x) = 1 nd the value of x in the
function
xxf
3
:
→
2
2
−
116
117
Present your work to class.
Now look at the given example below, compare your way of
working.
Example 3:
a) If
()
32
−= xxf
nd x if
()
5
=xf
this means
x
x
x
=
=
−=
4
28
32
5
b) Given g(x) = 3x +2 nd the domain given the range is {2, 11,
−1, −13} Solve equations, equating g(x) = 2, 11, −1, −13 in
turns.
0
30
232
=
=
+=
x
x
x
3
93
2311
=
=
+=
x
x
x
1
33
123
−=
−=
−=+
x
x
x
Therefore the range is
{}
5
,
1
,
3
,
0
−−
Exercise 6d
1. If
()
,
15
+= xxf
nd x if
a)
()
11
=xf
b)
()
6
=xf
c)
()
21
=xf
d)
()
14
−=xf
e)
()
.
49
−=xf
�
�
118
119
2. If
()
,
37
−= xxg
nd x if
a)
()
11
=xg
b)
()
3
−=xg
c)
()
137
=xg
d)
()
17
−=xg
e)
()
73
−=xg
3. If
()
5
24
−
=
x
xh
, nd x if
a) h(x) = 2
b) h(x) = − 2
c) h(x) = 7
d) h(x) = 10
4 If f(x) = x
2
+2x – 3, nd x if
a) f(x) = 21
b) f(x) = −3
c) f(x) = 0
5 If g: x → x
2
– 1, nd x if
a) g(x) = 3
b) g(x) = 8
c) g(x) = 1.25
6. If h(x) = x + 3, nd x if
a) h(x) = 5
b) h(x) = 0
c) h(x) = 4.5
118
119
Solving real life involving functions
In everyday life you unknowingly or knowing use relations in
business and other areas.
Activity 4:
Identifying real life functions
In groups discuss; which of the following could be functions.
State your reason.
(a) Petrol used by a car and distance travelled.
(b) Number of days and food used in the boarding school.
(c) Number of people and food used.
(d) Radius of a circle and area of the circle.
(e) Number of days spent in the boarding school and food left.
(f) Distance travelled and time taken in hours.
(g) Body mass and height of the person.
Report your answers to class in a plenary session and now look
at the example below.
Example 4:
Functions in real life
If an orange costs K10, how many oranges will one buy with
K30, K50, K80, and K100?
Solution
To get the number of oranges, divide the given amount by
the cost.
Let the function be f(x) = x/10, the domain is {30, 50, 80,
100}
Then f(30), f(50), f(80), f(100)
120
121
In table form, you have;
x 30 50 80 100
f(x) 3 5 8 10
Exercise 6e
1. In order to mitigate the impact of climate change, a village
raised tree seedlings of which only half planted became fully
grown. Let x represent the number of seedlings planted and
y represent the number that become full grown trees.
(a) Write an equation that shows the relationship described
above.
(b) Find four solutions of the equations. Write the solutions as
ordered pair.
(c) Why does negative value not make sense?
2. The table below shows the tax rates provided by the MRA
on salary earned.
Taxable income on monthly
income(MK)
Rate of tax
First 20,000 0%
Next 5,000 15%
Excess of 25,000 30%
Three people gets the following; K15, 000, K22, 000 and
K24, 000 as their monthly incomes.
Find how much will each pay to MRA. Give your answer as
an ordered pair.
3. Grace is saving money to be used to buy school bag for
K2200. She already has K1500 in her savings account. She
plans to add K80 each day from the money she earns for
selling freezes. The equation f(x) =1500 + 80x describes total
Grace’s total savings f(x) after x weeks.
After how many weeks will she have enough money to
purchase the bag?
120
121
4. The temperature in degrees of the mine varies with depth.
The temperature in degrees Fahrenheit, y is estimated by
y = 18x + 66.5 where x is the depth in kilometres.
(a) Create a table of ve ordered pairs of values that relate the
depth of mine and the temperature of its walls.
(b) What would be the temperature of walls be if the mine is
3.6km deep.
5. The formula for nding the perimeter of a square with s
units long is P = 4s. Find ve ordered pairs of values that
satisfy this condition.
Unit summary
• In this unit you have looked at functions. A function is rule
of correspondence between two sets such that exactly one
element in the second set corresponds to each element in
the rst set. The set of values in the domain is called the
input and a range of values is called a output. The next unit
looks at circle geometry where you will look at the angle
properties.
Unit review exercise
1. Let g(x) = √x and h(x) = 7 +2x
a) g(9)
b) h(9)
c) g(1)
d) h(1)
2. If f(x) = x
2
−2x, nd
a) f(1),
b) f(2),
c) f(4)
122
123
3. If f(x) =
3
24 5
,
nd
a) f(3)
b) f(−8)
c) f(−5)
4. For the function f: x → x
2
− 3, nd x if
a) f (x) = −2
b) f (x) = 22
c) f (x) = 13
d) f (x) = −3
5. Given that f(x) = 9
x
, calculate
f(
3
2
)
6. The gure below an arrow diagram for the function
f(x) = 4x − 2.
x f f(x)
3 9
5 q
p 6
Find the values of values of p and q.
7. Given that f(x) =
3
8
,
nd f (-b) in its simplest form.
8. Given that g(x) =
3
+1
, calculate the value of x when g(x) = 2.
9. The function g(x) =
2 1
is dened on the domain {−1,
1
2
, 1}.
Draw the arrow diagram to represent this function.
10. Given that f(x) = ax − 6 and f (6) = 18, nd a.
11. Translate the sentence to equation and nd four solutions
as ordered pair. Some number is 4 more than the second
122
123
number.
12. Linda buys drinks for a morning meeting. She knows that
the staff prefers fanta over coke. If she buys at least twice
as many fanta as cocoa, write a relation to show different
possibilities.
13. The table below shows the distance covered by a motorist
from from Lilongwe to Blantyre.
t(h) 0 1 3 5 6
d (km) 0 50 150 250 300
What type of relation is this?
Draw distance time graph. Find the time travelled in 4hours
time.
14. The table below shows approximate populations of certain
local towns.
Town A B C D E
Population 1600 3000 5000 10000 12000
If HIV prevalence rate in Malawi as of 2014 is at 10.5%.
If f(x) = 0.105x, calculate the number of people who are
infected in each town, assuming that the prevalence rate is
the same in the entire town. Give your answer as an ordered
pair.
15. As a thunderstorm approaches, you see lighting as it occurs,
but you hear the accompanying sound of thunder a short
time afterwards. The distance y in miles that sound travels
in x seconds is given by the equation y = 0.21x.
(a) Create a table of ve ordered pairs of values that relate
the time it takes to hear thunder and the distance from the
lighting.
(b) How far away is lighting when the thunder is heard 2.5s
after the light is seen?
124
125
Summary
The unit has looked at functions. A function is rule of
correspondence between two sets such that exactly one element
in the second set corresponds to each element in the rst set.
The next unit looks at circle geometry where you will look at the
angle properties.
Glossary
Function (or mapping) is rule of correspondence between two
sets such that exactly one element in the second set corresponds
to each element in the rst set.
Domain is the rst set of inputs of a function.
Range is the set of images of all the elements in the domain of a
function.
References
Fawdry J. B. (1994), Additional Mathematics a course of students,
Mathematics Association of Malawi.
Hau S. and Saiti F. (2010). Strides in Mathematics 3. Blantyre:
Longman Malawi.
Elaine R. et al, CHANCO Teach yourself series, Mathematics
Questions and Model answers, Second Edition (2013), Chancellor
College Publications, Zomba, Malawi
Thomo F. et al (2011). Excel and Succeed secondary
Mathematics for 3. Nairobi: Longhorn Publishers.
124
125
Unit
7
CIRCLE GEOMETRY
ANGLE PROPERTIES
In unit 3, you learnt about circle
geometry in relation to parts
of the circle, and properties of a
chord. In this unit, you will learn
about angle property in which
you will learn about the properties
of a circle and prove theorems
involving angles. You will
describe the properties of cyclic
quadrilateral and apply properties
of a cyclic quadrilateral to solve
problems.
The knowledge of circles is used in
different elds such as production
of wheels and tires, dinner plate
and coins in industries.
Describing angle
properties of a circle
You will begin the unit by rst
establishing the relationship between
the angle at the centre and that at
the circumference.
Activity 1:
Establishing that the angle at the
centre is twice the angle at the
circumference
In pairs Draw a circle of a reasonable
radius with centre O as shown below.
(a) Draw angle AOC with A and C
on the circumference.
(b) Draw another angle ABC
with B at any point on the
circumference.
(c) Measure angle AOB and angle
ABC
(d) What conclusion can you draw
on angle AOB and angle ABC?
Present your work to class.
A
B
O
C
Figure 7.1
126
127
You might have noted that the angle at the centre is twice the angle at the
circumference. I.e. angle AOB = 2angle ABC.
Can you try to show this by proving using the diagram above? Now compare
your work the with the proof shown below.
Theorem:
Angle which an arc of a circle subtends at the centre is twice the
angle which it subtends at any point on the remaining part of
the circumference.
Using the drawings shown below
(a)
A
P
O
B
E
H
(b)
A
P
B
E
H
(c)
A
P
O
B
E
H
Given: a minor arc AHB of the circle, centre O, and a point P on
the remaining part of the circumference
To proof: ∠AOB = 2∠APB
Construction: join PO and produce it to any point E
Proof: AO = OP (radii)
Figure 7.2
126
127
∠OAP = ∠OPA (base ∠s of isos.)
But ∠EOA is an exterior angle of AOP,
∠EOA = ∠OAP + ∠OPA (2 opp. int. ∠s of equal to
ext. ∠)
∠EOA = 2∠OPA
Similarly, ∠EOB = 2 ∠OPB
But ∠APB = ∠OPA + ∠OPA
And also ∠AOB = ∠EOA + ∠EOB
= 2∠OPA + 2∠OPA
∠AOB = 2APB
In pairs, using gure 7.3 below show that ∠APB = ∠AOB
consider the reex ∠AOB.
A
P
B
H
O
Now you will look at the angle in a semicircle.
Activity 2:
Establishing that the angle in a semicircle is a right angle
In pairs:Using a pair of compass and ruler, construct a circle of
reasonable radius with centre O.
(a) Draw a diameter AB
(b) Mark point C at any point on the circumference on opposite
side of the diameter
(c) Draw angle ACB and measure its size.
Figure 7.3
128
129
(d) What do you notice? Find out from your friends what they
have come up with. Well, you might have found out that an
angle in a semicircle is a right angle.
How can you prove that? Discuss with a friend if you can also
show the proof.
Theorem: The angle in a semi - circle is a right angle.
Given: a circle with centre O and a diameter AB subtending
∠ACB at the circumference.
A
B
O
C
To prove: ∠ACB = 90
Proof: ∠AOB = 2∠ACB (∠ at the centre twice the ∠
at the circumference)
But ∠AOB = 180
0
(∠ on a straight line)
2 ∠ACB = 180
0
∠ACB = 90
0
(divide both sides of by 2)
Thus angle in a semicircle is a right angle.
What is the relationship between the angles are in the same
segment which are subtended by the same arc or chord.
Activity 3:
Proving that angles subtended by the same arc/chord are
equal.
In groups discuss using the gure 7.5 below, show that a = b
Figure 7.4
128
129
(a)
O
A
B
P
M
a
b
(b)
O
A
P
M
B
Present your work to the class.
Now look at the theorem.
Theorem: angles subtended by the same arc / chord are equal
or angles in the same segment of a circle are equal.
A
P
M
B
O
Given: a circle with centre O and ∠APB and ∠AMB on
the circumference
To prove: ∠APB = ∠AMB
Construction: In (a) join AO and BO
Proof: ∠AOB = 2∠APB (∠ at centre twice ∠
at circ.)
Also ∠AOB = 2∠AMB (as above)
∠APB = ∠AMB
Similarly in (b) both angles are angles in a semicircle.
You will have to apply these theorems to solve some problems.
Here are some examples.
Example 1
In gure below, O is the centre of circle ABCD.
Figure 7.4
Figure 7.5
Figure 7.6
130
131
A
B
C
D
O
34
0
If ∠BAD = 34
0
, nd ∠BOD and ∠Bbncgn CD
Solution;
∠BOD = 2∠BAD = 2 x 34
0
= 68
0
(∠ at the centre = twice ∠
at the circum.)
Reex ∠BOD = 360
0
− 68
0 =
292
0
(∠at the point)
∠BCD = ½∠BOD = ½(292
0
) = 146
0
(∠ at the centre
= twice ∠ at the
circum.)
In gure 7.7, RT is a diameter of circle RSTV, centre O.
R
S
T
V
65
0
If ∠RTV = 65
0
, nd ∠TVS.
Solution;
In ∆TRS
∠RST = 90
0
(∠ in a semicircle)
∠TRS = 180
0
– (90
0
+ 65
0
) (∠ sum of)
= 25
0
But ∠TRS = ∠TVS (∠s subtended by the same
chord TS)
Figure 7.7
130
131
∠TVS = 25
0
Now do the exercise below.
Exercise 7a
1. Find the lettered angles in the given circles.
(a)
32
x
y
O
(b)
119
0
y
v
(c)
x
y
36
0
O
(d)
40
0
O
x
y
(e)
O
x
140
0
(f)
x
y
v
47
0
Cyclic quadrilaterals
Properties of a cyclic quadrilateral
Now you will look at cyclic quadrilateral. Can you dene a cyclic
quadrilateral?
A cyclic quadrilateral is a quadrilateral in which all the four
132
133
vertices lie on the circumference.
Activity 4:
Describing properties of a cyclic quadrilateral
In pairs;
(a) Draw a circle with centre O with reasonable radius as like
the one below.
(b) In the circle draw a cyclic quadrilateral ABCD.
(c) Using a protractor, measure all the angles of the cyclic
quadrilateral ABCD.
What can you say about the opposite angles of a cyclic
quadrilateral?
O
A
B
C
D
Now using the diagram above and your previous knowledge,
show that ∠A + ∠D = 180
0
.
Report your ndings to the class.
Did you note that the opposite sides are supplementary? The
sum of supplementary angle is 180
0
. Now look at this theorem.
Theorem: the opposite angles of a cyclic quadrilateral
are supplementary or angles in opposite segments are
supplementary
Given: a cyclic quadrilateral ABCD
To prove: ∠BAD + ∠BCD = 180
0
Figure 7.8
132
133
A
B
C
D
2y
y
2x
x
Join:
Join OB and OD
Proof: with letters in the gure above
∠BOD = 2y (∠ at the centre = twice ∠ at the circum.)
Reex ∠BOD = 2x (as above)
2x + 2y = 360
0
(∠s at a point)
x + y = 180
0
divide by 2 both sides
Hence ∠BAD + ∠BCD = 180
0
Having looked at that, now you will look at the interior and
exterior angle of a cyclic quadrilateral
Activity 5:
The interior and external angle of a cyclic quadrilateral
In groups
Study the gure below.
Figure 7.8
Figure 7.9
134
135
A
B
C
D
(a) Draw a similar gure and measure ∠ABC and ∠CDE.
(b) What do you notice?
(c) What is the relationship between these two angles?
(d) Identify the interior angle and exterior angle between ∠ABC
and ∠CDE.
(e) Discuss and show theoretically that ∠ABC = ∠CDE.
(f) Present your work to the class.
You might have noted in the gure above that ∠ABC = ∠CDE.
Now look at the theorem below.
Theorem: The exterior angle of cyclic quadrilateral is equal to
the interior opposite angle.
Given: a cyclic quadrilateral PQRS with PS extended
to T
To prove: ∠PQR = ∠RST
Proof: with letters in the gure
y + x = 180
0
(opp. ∠s of cyclic quad,)
y + v = 180
0
(∠s on str. line)
x = v (= 180
0
– y)
∠PQR = ∠RST
Now you can apply the knowledge acquired and solve some
problems. Look at the example below.
Figure 7.10
E
134
135
Example 2
In the gure A, B, C, D are points on a circle centre O. BA is
produced to E. If ∠DAE = 76
0
and ∠ADO = 69
0
, nd ∠ABO.
Solution;
∠BCD = 76
0
(= ext. ∠ of cycl. quad.)
∠BOD = 152
0
∠BAD = 180
0
− 76
0
(∠s on str. line)
= 104
0
Now in quadrilateral. ABCD
∠ABO = 360
0
−152
0
– 104
0
−69
0
(∠ sum of quad.)
= 35
0
∠ABO = 35
0
In the gure 7.12 below, CE is a diameter of a circle ABCDE. If
∠ABC = 126
0
, nd ∠ACE.
Figure 7.10
B
A
E
O
76
0
69
D
o
c
Figure 7.11
136
137
126
0
A
E
B
C
D
O
Solution;
∠ADC = 180 − 126
0
(opp. ∠ of cycl. quad.)
= 54
0
∠EDC = 90
0
(∠ in semicircle)
∠ADE = 90
0
– 54
0
= 36
0
But ∠ADE = ∠ACE (∠s in the same segment)
∠ACE = 36
0
Now do the following exercise.
Exercise 7b
1. Find the value of the variables
(a)
102
0
x
0
70
0
y
0
(b)
100
0
85
0
32
0
e
Figure 7.12
136
137
(c)
47
0
(d)
98
0
71
0
a
0
b
0
c
0
(e)
82
0
x
0
y0
z
0
110
0
(f)
105
0
f
0
g
0
115
0
2.In the gure 7.13 below, if DC is a diameter and O is the centre
of the circle, calculate angles BDC and DAB.
O
D
C
A
B
Now you will look at concyclic points.
Concyclic points and cyclic quadrilaterals
What do you understand by concyclic points? In last two
theorems, you were looking at cyclic quadrilaterals. Do
remember how you dened cyclic quadrilateral? Deduce the
Figure 7.12
Figure 7.13
138
139
meaning of concyclic points from there.
Concyclic points are points which lie on the circumference of a
circle.
Showing that points are concyclic
If the four points A, B, C, and D lie on a circle, then they are
said to be concyclic
Activity 6:
Showing that points are concyclic
In groups;
(a) What are the properties of angles in cyclic quadrilaterals?
(b) Discuss three ways on how you can show that points are
concyclic.
(c) Present your work to the class.
Now to show that points are concyclic points, you look at the
converse of the theorems you looked earlier in this unit.
The following holds true for concyclic points
1.If the angles subtended by the same line are equal for example,
angles APB, AQB, ARB are equal and subtended by AB.
Q
A
B
P
R
Which points are concyclic? Here ∠ APB = ∠AQB = ∠ARB and
A, P, Q, R and B are concyclic points.
If the opposite angles of a quadrilateral are supplementary, then
Figure 7.13
138
139
the quadrilateral is cyclic.
x
1
x
2
A
B
C
D
2. If x
1
+ y = 180
0
, then this shows that ABCD is a cyclic
quadrilateral and that A, B, C, D are concyclic points.
3. If the exterior angle of a quadrilateral is equal to the interior
opposite angle, then the quadrilateral is a cyclic.
From the gure above if x
1
= x
2
then ABCD is a cyclic
quadrilateral and A, B, C, D are concyclic points.
Now look at the example below.
Example 3
If ABCD is a quadrilateral in which ∠ABC = 95
0
, ∠BAC = 53
0
,
∠ADB = 32
0
.Prove that ABCD is a quadrilateral.
Solution;
Sketch the quadrilateral
32
0
A
B
C
D
53
0
95
0
In ∆ ABC
∠ACB = 180
0
– 95
0
– 53
0
(∠ sum in )
= 32
0
But ∠ACB =∠ADB = 32
0
(as shown)
Figure 7.13
y
140
141
Also ∠ACB and ∠ADB are subtended the line AB
ABCD is a cyclic quadrilateral.
Exercise 7c
1. ABC is an equilateral triangle and ACD is an isosceles
triangle drawn outside ∠ABC such that DA = DC and ∠DCB
is a right angle. Prove that A, B, C, D are concyclic.
2. PQRS is a trapezium having PQ parallel to SR and ∠PSR =
∠QRS = 73
0
. Prove that PQRS is a cyclic quadrilateral.
3.In the gure below, O is the centre of the circle ABP, MO is
perpendicular to AB, and BPM is a straight line.
A
O
B
L
M
P
Prove that
(a) A, O, P, M are concyclic points.
(b) Angle OPA = angle OMB
4. In the gure below, ABCD is a cyclic quadrilateral and O is the
centre of the circle. AB is parallel to DC and BC produced meets
AD produced at M.
140
141
C
O
A
B
D
M
Prove that;
(a) MCD is isosceles triangle
(b) ODMB is a cyclic quadrilateral
5. MNYX is a circle centre O in which NM = YX. When MN and
XY are produced, they meet at P. the mid – points of NM and
YX are E and F respectively.
Prove that;
(a) Triangles OEP and OFP are congruent
(b) Points O, E, P, F are concyclic.
Unit summary
• The unit has so far covered work on properties of a cyclic
quadrilateral where a number of theorems have been
discussed and illustrated.
• The unit has nalized by looking at concyclic points. It has
exercises for practice. The next unit looks at transformation.
142
143
Unit review exercise
1. The gure is a cyclic quadrilateral ABCD where AD = AB.
The diagonal BD = BC.
52
0
A
B
C
D
If ∠BDC = 52
0
, nd ∠ABC.
2. ABCD is a circle and the chords AC, BD cut at X. P and Q
are points on XC, XD respectively such that PQ is parallel to
CD. Prove that ABPQ is a cyclic quadrilateral.
3. Show that the gure ABCD below is a cyclic quadrilateral.
65
0
75
0
40
0
A
B
C
D
4. ABC is an equilateral triangle. A line from A meets BC at R,
another line from B meets CA at S such that BR = CS. The
two lines BS and AR, intersect at Q. Prove that;
(a) Triangles ABR and BCS are congruent.
(b) Quadrilateral RCSQ is cyclic.
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143
5. The gure ABCD below is a circle, centre O. AOC is a
straight line and AB = BD.
O
A
C
B
D
If angle ABD = 70
0
, calculate angle BAC.
6. A, B, C and D are points on a circle below with BOC a
diameter, AD = CD and ∠DBC = 40
0
.
O
D
C
40
0
A
B
a
c
d
b
Find the values of the angles marked a, b, and c.
Glossary
A cyclic quadrilateral is a quadrilateral in which all the four
vertices lie on the circumference.
Concyclic points are points which lie on the circumference of a
circle.
References
Geoff Buck well.(1997) Mastering Mathematics. Macmillan.
London.
Channon J. B. et al. New General Mathematics 3. A modern
Course for Zimbabwe. Longman. London.
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145
Ric Pimentel and Terry Wall.(2011) IGCSE Mathematics 2
nd
Edition.John Murray. UK
Elaine Ryder et al.(2013) CHANCO MSCE Mathematics
Questions and Model Solutions. Chancellor College. Zomba.
Gunsaru et al. Secondary Mathematics Book 3. Dzuka
Publishing Company. Blantyre. Malawi.
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145
Unit
8
TRANSFORMATION
Recall that in your JCE course,
transformation was dened as the
change in shape or position of an object.
You studied reections and rotations
as examples of transformation. In this
unit you will learn about more types
of transformation: translation and
enlargement. You will learn how
to rotate a simple plane gure about
a given point through a given angle
clockwise or anticlockwise. You will also
learn to translate a simple plane gure
in a column vector. Finally, you will
learn to enlarge a simple plane gure
by a scale factor and a given a centre of
enlargement.
The knowledge of transformations is
used in artistic designs to make objects
more appealing. People like, doctors and
surveyors use enlargement in their day
to day work.
Names and labels in
transformation
Throughout this unit you will
use the words object to mean the
original gure before transformation
and image to mean the gure after a
transformation. All corresponding or
matching points on the object and the
image will have to be named using
the same letter or same numbers.
Where letters are used, those on the
image should have a prime ´ on them.
Where numbers are used, they should
appear as subscripts.
Rotating a plane gure
about a given point
A rotation is a turn. In a rotation
an object turns about a xed point
called centre of rotation. A rotation
can be clockwise or anticlockwise.
When the rotation is clockwise it is
negative while if it is anticlockwise,
it is positive. The rotation is given
as a fraction of a turn or an angle in
degrees.
Activity 1:
Drawing rotations
In groups
1. Draw a grid of square boxes
(10 boxes by 10boxes) on a
paper. You can also use a
squared paper if you have it.
2. Using a cardboard paper or
any other hard paper, cut out a
right angled triangle and x a
stick to one corner as shown on
following page:
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147
3. Pivot one end of the stick at one point on the drawn square
grid as shown below:
Pivot here
4. Now rotate the triangle through angles 30
0
and 45
0
clockwise, each time drawing the image of the triangle.
Measure the angles that each vertex rotates through at the
pivot centre.
5. Now draw square grids of your choice. Using different plane
shapes e.g rectangles, trapeziums and your own choices of
angles and directions, draw rotations of the plane shapes on
the square grids. Again measure the angles that the vertices
of your diagrams go through at the pivot centre.
6. Comment on your ndings.
In a rotation, all points on the object move by the same
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147
measurement. To draw a rotation you must be given the
centre of rotation, the angle of rotation and the direction of
rotation. Here are two ways of drawing rotations:
Using a protractor and a ruler:
The assumption is that the initial position of the plane shape is
given or the gure is drawn on the grid or squared paper.
• Join each vertex of the plane shape to the given centre of
rotation.
• Choose any one vertex and using a protractor, draw the new
position of this line by measuring the angle of rotation.
• Measure the distance of the vertex from the centre of
rotation and use this distance to locate the new position of
the vertex on the new position of the line. Do the same with
other vertices.
• Join the new positions of the vertices on the new lines to
obtain the image of the plane shape.
Using tracing transparent paper
• Join one point on the gure to the centre of rotation.
• Place the tracing paper on the squared paper on which the
gure is drawn.
• Trace the shape, the line drawn in the rst step above and
the centre of rotation on the tracing paper.
• Measure the given angle from the line drawn in the rst
step.
• Place a sharp point say a pencil or pen or compasses on the
traced centre and actual centre.
• Turn the traced gure through the given angle until the line
drawn in the rst step coincides with the line or mark you
made when you measured the angle in step 4.
• Finally, trace the gure to obtain the new position of the
object.
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149
Exercise 8a
1. Use the above information and the grid below to draw the
image of drawn triangle
a) after a rotation of 90
0
clockwise about (0,0)
b) after a rotation of 45
0
anticlockwise about the origin.
c) after a rotation of -180
0
anticlockwise about (0,0)
d) after a rotation of -270
0
clockwise about (0,0)
=ñ=
= == =
= =
= =
= =
= =
= =
= =
= =
= =
= =
= =
= =
= =
2. On a grid draw rectangle with vertices at (-2,1),(-6,1) (-6,3)
and ( -2,3). Draw the image of the rectangle after a rotation
of 45
0
about (0, 0).
3. Draw and number x- and y- axes from -8 to +8. Show
the position of a point Q (4,3) after each of the following
rotations. In each case state the coordinates of the new
point.
a. 180
0
about (0,0)
b. +90 about (1,4)
c. +90
0
about (-6,5)
d. -90
0
about (-3,0)
4. Trace the shape below. Find the image of the shape after a
rotation +120
0
about X.
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149
X
Activity 2:
Describing a rotation
To describe a rotation you must give the angle of rotation, the
centre of rotation and the angle of rotation whether clockwise or
anticlockwise.
Your teacher will provide you with a gure and its image on a
squared grid. The centre of rotation will also be given. In groups,
1. Join one vertex on the object to the centre of rotation.
2. Join the corresponding vertex on the image to the centre of
rotation.
3. Using a protractor, measure the angle between the lines
joining the vertices to the centre of rotation.
4. Describe the rotation.
Sometimes you may not be given the centre of rotation. When
this happens, you will have to proceed in the following way:
1. Join two pairs of matching points on the object and the
image.
2. Using a pair of compasses, draw the perpendicular bisectors
of the two lines joining the points.
3. Extend the lines until they intersect at a point. This point
gives the centre of rotation.
4. To nd the angle of rotation, join any one pair of matching
points on the object and the image to the centre of rotation.
Measure this angle using a protractor.
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151
Example 1:
Describing rotation
(a) Describe the rotation mapping triangle A onto B in each of
the following diagram. C is the centre of rotation and the
shaded triangle is the object.
=
=
=
=
=
=
=
=
=
=
=
=
=
_=
C
^
Solution
The rotation is 90
0
clockwise about centre C or you may say that
the rotation is 270
0
anticlockwise about centre C. If you want to
use – or + you can write –90
0
about C or +270
0
about C.
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151
(b) Describe the rotation that maps arrow A onto A´
y
x
=
==
^
=
=
==
==
-3 JO===================== =JN==================== =
=
M====================
=
==N============ ============O===
ñ
A′
N=
2
-1
JO=====================
Solution
Join the matching points on the two arrows and construct the
perpendicular bisectors of the lines.
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153
=
==
^
=
=
==
==
-3 JO===================== =JN==================== =
=
M====================
=
==N============ ============O===
ñ
A’
N=
2
-1
JO=====================
Lines joining matching points to the
centre of rotation. The angle between
them is the angle of rotation.
(Dashed lines are perpendicular bisectors of lines joining
matching points on the object and the image)
From the diagram, the rotation is 90
0
clockwise about (1,-2).
Exercise 8b:
1. The vertices of a triangle ABC are (3, 3), (1,-1) and (3,-2).
The vertices of the image of triangle ABC are (-3, 3), (1, 1)
and (2, 3) respectively. Graph the two triangles on a squared
paper and describe the rotation mapping triangle ABC onto
its image.
2. The vertices of a triangle ABC are (-1,1), (-2,1) and (-2.5,2) .
x
y
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153
The vertices of the image of triangle ABC are (5,1), (6,1) and
(-5.5,0) respectively. Graph the two triangles on a squared
paper and describe the rotation mapping triangle ABC onto
its image.
3.
4
3
P 2
C
1 Q
-4 -3 -2 -1 0 1
In the above diagram, P is rotated anticlockwise about C to
obtain imge Q. Describe the rotation.
4. A quadrilateral WXYZ has coordinates (-4,0),(-3,1),(-2,0)
and ((-2,0). The Image of the quadrilateral WXYZ are (0,-
2),(-1,-3),(-2,-3) and (-2,-2). Draw the two quadrilaterals on
a squared paper or grid and nd the centre and angle of
rotation.
5. Triangle ABC and PQR are congruent in that order.
Triangle ABC has Vertices (-4, 3), (-3.8, 0) and (-2.5, 1).
Triangle PQR has vertices (0.1,-2.5), (1.9, 0) and (0.2, 0.1).
By drawing grids or by using a squared paper, describe the
rotation that maps triangle ABC onto triangle PQR.
Translation
A translation sometimes called a slide or a shift moves the
shape in a straight line. The shape of the object does not change
and every point on the object moves by the same amount and in
the same direction.
An example of a translation is shown below:
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155
Object
Image
Activity 3:
Describing translation
Your teacher will provide you with a graph paper. In your
groups,
1. Using a scale of 2cm to represent 1 unit on both axes, draw
a triangle PQR such that the vertices are (-3, 3), (-2, 5) and
(-5, 0) respectively.
2. On the same axes draw the image of triangle PQR such that
the vertices are (0,-1), (1, 1) and (-2,-4) respectively.
3. Use your drawings to describe the translation mapping
triangle PQR onto P΄Q΄R΄.
To describe a translation you must give the distance and the
direction an object has moved from its old position to its new
position. The movement is given in two parts: horizontal and
vertical. You start by giving the movement in the horizontal
direction followed by the movement in the vertical direction.
Example 2:
Translation
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155
-6 -5 -4-3 -2 -1 0 1 2 3 4 5 6 7 8
5
4
3
2
1
-1
-2
A
B
y
x
Describe the translation that maps rectangle A onto rectangle B
in the grid above.
Solution:
As all points move by the same amount in a translation, you
choose any one point on the object and trace the units the point
moves to the corresponding point on the image. For example if
you choose the top left corner, you will see that the rectangle
moved 10 units horizontally to the right and then 8 units
vertically downwards.
A
B
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157
Exercise 8c
E
A F
H
C
K
D
G
B
L
J
M
P
N
1. Using the above grid, describe the translation that moves
the following points: Each box is 1 unit by 1 unit long. For
parts k to 0, give a single translation.
a. A to B
b. C to F
c. D to Q
d. N to K
e. H to G
f. J to E
g. L to M
h. A to Q
i. F to K
j. E to M
k. C to N.
l. B to C to H
m. Q to J to D
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157
2.
A B
C
C�
B
�
A
�
In the above grid each grid is 1 unit by 1 unit .Describe the
translation that maps each gure onto its image.
Activity 4:
Drawing a translation
To draw a translation, you must be given the amount of
movement of the object both horizontally and vertically. You
must also be given the position of the gure to be translated.
Your teacher will show you a chart on which different plane
shapes are drawn. In pairs, draw the grid and copy the gures.
Translate the gures and draw the image of each gure as
follows:
1. Translate A, 3units right 4units down to A
1.
2. Translate B, 5units left 2units up to B
1
3. Translate C, 4 units down to C
1
.
Exchange your work with your partners and mark each others’
work. Let your teacher check your work.
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159
Exercise 8d
A
B
C
D
Draw the image of each of the above plane shapes after the
following translations:
1. Shape A:
(a) 4 units to the right and 2 units downwards.
(b) 10 units to the right.
2. Shape B:
5 units down.
3. Shape C:
(a) 6 units to the right.
(b) 3 units downwards.
4. Shape D:
7units to the left and 9 units upwards.
Translation and column vectors
In Form 2, you learnt that vectors represent movements. In
pairs discuss the meanings, in terms of movement, of the
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159
following:
, , , . These are known as translation
vectors. The translation vector shows how much the image has
been moved in relation to the object. It is written in the form
, where x and y are real numbers. x represents the units in x
axis, and y in y axis. When x is negative, movement is to the left
while if it is positive, movement is to the right. Similarly, when
y is negative, movement is downwards while if it is positive,
movement is upwards.
Activity 5:
Writing down the coordinates of a translation in column
vectors
In pairs, discuss how you can write the following coordinates of
translation in column vectors basing on the paragraph above:
a. a movement of 2 units to the left followed by a movement of
1 unit upwards.
b. a movement of 3 units downwards.
c. a movement of 4 units to the right followed by a movement
of 5 units downwards.
Present your work to class.
Example 3:
Writing coordinates
Write the following coordinates of translation in column vectors:
(a) A movement of 1 unit to the left followed by a movement of
2 units downwards.
(b) A movement of 5units upwards.
Solutions
(a)
(b)
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161
Exercise 8e
Write the following coordinates of translation in column vectors:
A movement of
1. 2 units to the left followed 5 units down
2. 5 units to the right followed by 2 units up.
3. 5 units to the left.
4. 6 units to the right followed by 1 unit down.
5. 1 unit up.
6. 8 units down.
7. 4 units to the right.
8. 2 units to the right followed by 8 units down.
9. 5 units to the right followed by 4 units down.
10. 7 units to the left followed by 7 units up.
Activity 6:
Translating shapes using column vectors
In pairs,
1. On a squared paper or on a grid, draw x- and y- axes and
number them from −5 to +5.
2. Draw triangle ABC such that A, B , C are points (-5,5),
( -2, -4) and ( 3,3) respectively.
3. Draw the image of triangle of triangle ABC after a
translation
.
4. Write down the coordinates of the vertices of image of the
triangle.
5. What is the relationship between the coordinates of the object,
the coordinates of the image and the translation vector?
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161
6. Compare your work with other groups.
To translate a plane shape using a column vector, you need to
rst describe the vector itself. To describe the vector means
to say what the vector means in terms of movement both
horizontally and vertically. Once the vector has been described,
the object can be moved by moving each vertex according to the
translation vector. When all the vertices have been moved, the
image can then be completed by joining the moved vertices.
You should also have seen from activity 6 that if the coordinates
of the vertices are written as vectors, the coordinates of the
object, the coordinates of the image and the translation vector
are related as follows:
Vectors from the coordinates of object vertices + translation
vector = Vectors of the coordinates of the image vertices.
Example 4:
Drawing images after translation
m
o n
-5 -4 -3 -2 -1 0 1 2
3
4
4
3
2
1
-
1
-2
-3
-4
Draw the image of triangle PQR after a translation of
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163
Solution
The vector means the object is moved 6 units to the right
and then 4 units downwards. The image is shown below. Dotted
lines have been included to show the movement of the object.
m
o
n
m´
-5 -4 -3 - 2 -1 0 1 2 3 4
o ´
n ´
y
4
3
2
1
-1
-2
-3
-4
Example 5:
Finding coordinates
A point R (3,-7) is translated to R´ by a translation vector T
Find the coordinates of R´.
Solution
Vectors from the object vertices + translation vector → Vectors
of the image vertices.
+ →
The coordinates of R´ are (-2,-11)
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163
Exercise 8f
1. In questions 1 to 3, copy the diagram and draw the image of
the object under the translation given by the column vector.
S
Q
P
O
N
M N O P Q R S T U
The vector is
R
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165
2.
JP JO JN M N O P Q R
The vector is
3.
Q
P
O
JQ JP JO JN M
JN
N O P Q
JO
JP
R
S
Q
P
O
N
N
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165
The vector is
4. A vector T translates a quadrilateral ABCD onto
quadrilateral A´B´C´D´.
(a) If A and B are points ( 1,3) and ( 3,4) , nd the coordinates of
A´ and B´.
(b) Given that the coordinates of C´ and D´ are (9,-3) and ( 7, -4)
nd the coordinates of C and D.
5. A translation vector T translates point A onto point A´. The
coordinates of A and A´ are (4, -2) and (7, 5). Find T.
6. A =
and B = are translation vectors. Find
the coordinates of a point P (3,2) after the Following
translations:
(a) A + B
(b) 2A – B
Enlargement
All the transformations you have studied so far (reections,
rotations and translations) have moved the object or turned
it over to produce the image, but its shape and size have not
changed. In each case, the image and the object are congruent.
In this section you shall learn about a transformation that
keeps the shape of the object but alters its size – enlargement.
Enlargement covers both making the image larger than the
object and making the object smaller than the object.
Finding the scale factor of an enlargement
In unit 12 of this book you dealt with scale factor. In that unit,
scale factor was dened as “ a number showing how many times
an object has been enlarged”. You also learnt that scale factor is
found by the formula
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167
Scale factor =
Example 6:
Finding a scale factor
Find the scale factor used in enlarging triangle ABC onto
triangle A´B´C´ in the above grid.
Solution
By counting the grids, AB = BC = 1 unit long and A´B´ = B´C´ =
4 units long. The scale factor can be found by dividing A´B´ by AB
or B´C´ by BC. You can also count the squares diagonally to nd
AC = 1 diagonal and A´C´ = 4 diagonals and then divide A´C´ by
AC.
Hence Scale factor =
= 4
Sometimes you may have to form right angled triangles and use
them to nd the scale factor as in example 7 below:
Length of the image
Corresponding length of the image
4
1
B
C
A
A
B
C
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167
Example 7:
Finding scale factor
m´
o ´
m
n ´
o
n
Find the scale factor used to enlarge triangle MNO to produce
triangle M´N´O´.
Solution
Choose any two corresponding vertices and draw lines
horizontally and vertically to produce right angled triangles as
follows (dashed lines have been used so that you can clearly see
how the triangles have been produced):
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169
m´
o ´
m
n ´
o
n
You can then divide the corresponding bases or heights of the
formed triangles to nd the scale factor.
Hence using bases, scale factor = = 2
Example 8:
Finding a scale factor
The coordinates of A and B are (2, 4) and (-3, 6) respectively. The
coordinates of A´ and B´ are (2.5, 5) and (-5, 8). Find the scale
factor used to enlarge AB to A´ B´.
Solution
You can use distance formula as follows:
Scale factor =
( 5 2.5)
2
+(8 5)
2
(
3 2)
2
+(6 4)
2
=
65 25
29
= 1.5 (from the calculator)
4
2
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169
Exercise 8g
Find the scale factor of enlargement in each of the diagrams
below( Questions 1 to 3).The shaded shape is the object.
1.
2.
4
2
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171
3.
4. The coordinates of the vertices of a triangle XYZ are (2, 3),
(4, 7) and (-1, -4).
The coordinates of the vertices of the image X´Y´Z´ of
triangle XYZ are (0.5, 4.5), (5.5, 14.5) and (-7,-13). Find the
enlargement scale factor.
5. A scale factor of
1
2
is used to enlarge a line segment with
end points P( -1, -5) and Q( 4,4). If the coordinates of Q´ are
(2, 2) and of P´ are ( x, -2.5), nd the negative value of x.
6. A rectangle 20 cm long and 15cm wide is enlarged by a scale
factor of 2. Find the length of the new rectangle.
7. A´B´C´ is the image of triangle ABC after an enlargement.
AB = 7cm, AC = 8cm, A´B´ = 14cm and B´C´ = 18cm.
(a) Find the scale factor of the enlargement.
(b) Find BC and A´C´
8. Two places are 4.25km apart and are presented on the map
by a distance of 18.5cm. Find the scale factor used.
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171
Finding centre of enlargement
The centre of enlargement is found by drawing straight lines
through matching points on the object and image. These lines
are then extended until they meet. The point at which they meet
is the centre of enlargement. Always draw three lines through
matching points. Two points give the point you want but the
third one acts as a check. See the diagram below.
O is the centre of enlargement
Exercise 8h
Find the centre of enlargement in each of the following diagrams
(questions 1- 4).
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173
y
0 5 10 15 x
10
5
0
2.
-5 0 5
x
5
-5
3.
1.
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173
y
0 5 10 x
5
4.
y
====== == = == =
====== == = == =
====== == = == =
====== == = == =
====== == = == =
====== == = == =
====== == = == =
= =
== =
=
== == === =
====== == = == =
(Each box is 1 unit long)
Drawing enlargement
In this section, you will learn to draw enlargement when given
a positive whole or negative scale factor. You will do this in two
ways:
(a) Drawing enlargement on a squared paper or grid when the
object and scale factor are given.
(b) Drawing enlargement on a squared paper or grid when the
object scale factor and centre of enlargement are given.
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175
Activity 7:
Drawing enlargement on a squared paper or grid when
the scale factor is given
In your groups and using grids of 1 unit long,
1. Draw a rectangle ABCD in which the length = 4 units and
the width = 2units.(Draw the rectangle parallel to the axes)
2. If you enlarge the rectangle by a scale factor of 2, how many
units would be the lengths and the width?
3. If you enlarge the rectangle by a scale factor of
, how many
units would be the lengths and the width?
4. Draw the images in 2 and 3 above on the same grid as the
rectangle ABCD above.
5. How do the images compare with the rectangle?
6. If the rectangle were drawn at an angle to the axes, discuss
how you would enlarge the rectangle.
When the lines forming the object run along the lines of the
grids, the lengths of image are found by just multiplying the
lengths of the object by the scale factor. If the lines are at an
angle to the lines of the grids, you need to nd the number of
grids “across” and “up or down” each line forming the object.
Multiply the number of the grids by the scale factor to nd the
lengths of the lines in the image.
In activity 7 you should also have seen that a positive whole
number scale factor makes the object larger than the original
object while a fractional scale factor reduces the size of the
object.
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175
Example 9
Enlarge triangle ABC above by the following scale factors
(a) 3
(b)
1
2
Solution
(a) For line AB there are 3 grids across and 4 grids up. A´B´ will
therefore be (3 × 3) = 9 grids across and (4 × 3) = 12 grids
up. Similarly, line B´C´ will be (4 × 3) = 12 grids across and
(2 × 3) = 6 grids up and line A´C´ will be (1 × 3) = 3 grids
across and (2 × 3) = 6 grids down.
(b) Multiplying the number of grids across and up each line by
1
2
as in (a) above, A´B´ will have 1.5 grids across and 2 grids up,
B´C´ will have 2 grids across and I grid up while A´C´ will have
0.5 grid across and 1 grid down.
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177
Activity 8:
Drawing enlargement on a squared paper or grid when
the scale factor and centre of enlargement are given
In groups you are to enlarge a triangle ABC by a scale factor of
3.
1. On a square paper or grid draw triangle ABC with vertices
in the corners of the grids. The triangle shouldn’t be too big.
2. Choose one point O and from that point, draw and extend a
straight line through each vertex of the triangle.
3. Using a pair of compasses or a ruler, measure the distance
from O to vertex A.
4. Multiply the distance in 3 by the given scale factor and from
O, mark off the new distance along the line OA and call the
marked point A´.
5. Repeat 2,3,4, for all the vertices and join A´,B´ and C´.
6. Measure the lengths of new triangle and compare them to
the matching lengths on the object. How do they compare?
7. Report your ndings.
176
177
To draw an enlargement, you follow the following steps:
1. From the centre of enlargement, draw and extend straight
line through each vertex of the object.
2. Using a pair of compass or a ruler, measure the distance
from the centre of enlargement to each vertex of the object.
3. Multiply the distance in 2 by the given scale factor and from
the centre of enlargement, mark off the new distance along
each line.
4. Join the last marks to obtain the image.
Example 10
The vertices of a triangle are A(4,2) , (6,4) and (3,6). Draw the
triangle on a grid of at least 17 by 17 boxes. On the same grid
draw the image of triangle using scale factor 3 and centre of
enlargement O (1,2). State the coordinates of the image.
Solution:
The coordinates of the image are (10,2) , (16,8) and ( 7,14)
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179
Exercise 8i
Copy the following diagrams and enlarge the objects by the scale
factor given and from the centre of enlargement shown. Grids
larger than those may be needed.
1.
Enlarge triangle ABC above using scale factor 3 and (0,0) as
centre of enlargement. What are the coordinates of the image?
178
179
2.
Draw the enlargement of the quadrilateral above using centre
(6,7) and scale factor of 4. State the coordinates of the image.
3.
Draw the enlargement of triangle PQR using (4,2) as centre
180
181
of enlargement and scale factor 2. Sate the coordinates of the
image.
4.
0
y
0000
o x
0 5 10
10
5
-
5
Enlarge the quadrilateral above by the scale factor using (8,5)
as centre of enlargement. State the coordinates of the image.
180
181
5.
Draw the enlargement of the quadrilateral above using (0,0)
as centre of enlargement and scale factor 2. What are the
coordinates of the image?
Unit summary
• In this topic you have learnt drawing rotations on a
squared paper, describing rotations using directions and
angles, describing and drawing translations, writing down
coordinates of a translation in column vectors, translating
shapes using column vectors, dening enlargement, nding
the scale factor and centre of enlargement and enlarging
shapes by a positive whole number scale factor and fractional
scale factor.
182
183
Unit review exercise
1. Draw, on a squared paper, a rectangle ABCD with
vertices at (-3,2),(-1,2), (-1,1) and (-3,1). Draw a rotation
of the triangle 90
0
clockwise about O (0,0). What are the
coordinates of the image of the image of the triangle?
2. T´(0,2) is the image of T(-1,-3) after a rotation about a xed
point C. Describe the rotation.
3.
y
X´(2,3)
x
X(-3,-2)
Describe the translation mapping point X onto X´ in the above
diagram.
4. The coordinates of the vertices of a triangle ABC are (3,1),
(1,2) and (1,5). Draw the image of the triangle after a
translation of
. What are the coordinates of the vertices
of the image?
5. A trapezium is moved 5 units to the left and 6 units down
from a point T. Write this translation as a column vector.
6. On a squared paper, draw triangle ABC with vertices
(3,5),(3,3) and (6,3) and the image of triangle ABC with
vertices (1,7), (1,1) and (10,1). Use the two drawings to nd
the centre and scale factor of enlargement.
182
183
Draw the enlargement of the body of a toy car shown above
using the black dot as the centre of enlargement and scale factor
2.
Glossary
Rotation: Transformation in which a gure changes the way it
is put.
Centre of rotation: A point about which a gure rotates .
Translation: Transformation in which a gure changes its
position.
Enlargement.: A transformation in which the gure changes
in size.
Centre of enlargement: A point about from which a gure is
enlarged.
Scale factor: The ratio of corresponding lengths of the image
and the object.
184
185
References:
Ralge Chikwakwa et al, Senior Secondary Mathematics Book3
(2002), Mamillan, Malawi.
Duncan and Christine Graham, Mainstream Mathematics for
GSCE(1996), Macmillan, London.
Suzanne et al, Middle Grades Math Course 3(1999), Prentice
Hall, United States of America.
184
185
Changing the subject of
literal equations
In most cases banks and other money
lending institution charge interest
on money lent to people and as they
such use formulae. Write down the
formula for calculating interest.
The formula for nding interest
is; Can you come up with
other formulae? Now do the activity
below.
Activity 1:
Identifying the subject of the
formula.
In pairs discuss the following:
1. Dene a formula.
2 Write the formula for nding the
following;
a. Perimeter of a rectangle
b. Volume of cylinder.
c. In these two formulae above,
which one is the subject.
d. How can you nd the length of
the rectangle
Present your answers to the class.
A formula is an equation that
shows a relationship between
two or more variables. Changing
or transposing the subject of a
formula is the same as solving for
an unknown or expressing the given
letter in terms of other letters. This
involves simple formula involving
literal equations and also others that
contain powers or roots.
Unit
9
CHANGE OF SUBJECT OF A
FORMULA
In your JCE Mathematics , you
learnt about linear equations.
You learnt about some basic
problems involving change of
subject of formula. In this unit,
you are going to learn more about
changing the formula of literal
equations and those involving
powers. Most often, formulae
can be used to solve any other
unknown in the formula provided
enough information is given.
The knowledge of change of
formula will help you to solve a
particular problem. For example
you can nd velocity of a car or
nd distance travelled given some
variables.
PTR
100
I=
186
187
Example 1:
Change of subject of formula
Make x the subject of the formula
(a) 3y = 4x − 5
Solution
Add 5 to both sides
3y + 5 = 4x − 5 + 5 … like terms together
3y + 5 = 4x
Divide both sides by 4
( )
33
×a
∴
x =
4
53
+y
(b) Make y subject of the formula x = 2(y + z)
Solution;
Divide both sides by 2
2
t
d
= y + z
Subtract z from both sides
2
t
d
− z = y + z − z
∴
2
t
d
− z = y
∴
y =
2
t
d
− z
186
187
Example 2:
Change of subject
of formula
Make R the subject of
the formula
100
PT
I
R
=
Multiply both
sides by 100
I × 100 = PTR
Divide both sides
by PT
PT
PT
PT
100 I R
=
×
PT
I 100
=∴R
Now do the exercise.
Exercise 9a
Make the letter
written in bold a
subject of the formula.
(a) A =
3
4
bh
b
(b)
1
+
1
=
1
u
(c) Y = kx
a
k
(d) x = b + d
d
(e) a = bd + f
b
(f) A =
3
4
(a + b)h
h
(g) V =
π
r
2
h
h
(h)
Q
Q - P
tr
=
Q
(i)
()
n 90 42 °×−= nS
All the examples
above involve literal
equations. These are
easy to express in
terms of any given
letter. However others
involve powers and
roots.
Formula involving powers
Some formula will
involve powers. From
exercise 9a you can
identify formulae that
have a power in it.
188
189
Activity 2.
Identifying the subject of the formula.
Discuss with a friend the following;
(a) What is the formula for nding area of a circle? Identify the
subject of the formula.
(b) Write the formula for nding the radius of the circle.
Now look at the following examples.
Example 3:
Change of subject of formula
Make A the subject of the formula Y = A
2
+ BD
Y = A
2
+ BD
Subtract BD from both sides
BD - Y
BD Y
=∴
=−
2
2
A
A
Find the square root of both sides
BD - Y
BD - Y
±=∴
=
A
A
2
Note: At Y − BD = A
2
, we cannot divide both sides by A
since both sides will have an A. When we make a
certain letter the subject of a formula, then that letter
should only be found on one side of the formula, and
not both sides.
Working with exponential and logarithmic equation
You will look at more example of that involve powers.
Example 4:
188
189
Given that
1 A
3
+= rh
(a) Make r the subject of the formula
(b) Find the value of r when
4 h and 10A ==
Solution
(a)
1 A
3
+= rh
Divide both sides by h
1
A
3
2
+=
r
h
Square both sides
1
A
3
2
+=
r
h
Subtract 1 from both sides
3
2
1 -
A
r=
h
Find the cube root of both sides
r =
3
2
2
1
A
−
h
For part b substitute the given values in the formula and
nd the value of r.
Subject of formula that involve exponential and logarithmic
functions
Some equation cannot be easily solved and such may require the
use of logarithm. See the example below.
Example 5
a. Make n the subject of the formula y = kx
n
190
191
Working out:
Introduce log on both sides
logy = logk + logx
n
Take logk to other side
logy – logk = logx
n
changing n from an exponent
nlogx = logy – logk
divide throughout by logx
therefore n =
Exercise 9b
In questions 1 − 18, a formula is given. A letter is printed in
bold after it. Make that letter the subject of the formula.
1. P
2
mNb +=
N
2. b = 2a
c
c
3. V
2
2 -
2
asu=
s
4.
y
x
m
π
=
y
5.
h A
32
rhr −=
π
6.
m
M
-M
mpn
m
x
+
=
7.
d
d
qrp
v
+
=
8.
()
[]
a 1-na
2
d
n
S +=
190
191
9. x =
s a
s
b
s
10.
a
2
b
A
222
bc
ac −+
=
11.
m P
5
nm
m
−
=
12.
h
3
2
V
+= rhr
π
13.
( )
q
222
qapq −=
14.
r K
2
rb
rtb
−
+
=
15.
A
4
2
r
π
=
r
16.
3
4
V
3
r
π
=
r
17. Y = kx
a
a
18. T = 2
l
Real life application of formula
Some of the problems solved in this unit involve real life
situation. Can you identify some of them from those given in
exercise above? Examples include;
Velocity, V = u + at
Volume of cylinder, V =
π
r
2
h
Area of a triangle; A = ½bh
There many more formulae that are used in everyday life.
192
193
Unit summary
You have learnt how to identify the subject of formula. You
also changed subject of formula of literal equations and those
involving powers.
In the next unit, you will learn about exponential and logarithm
equations.
Unit review exercise
1. Make y the subject of the formula in x = 2 (y + z)
2. Make p the subject of formula in the equation, logy = logx
p
+
log k
3. Make x the subject of the formula a
x
= b
4. Given that the area A of a triangle with base b and of height
h is equal to
.
2
1
A bh=
Make h the subject of the formula.
5. Given that
+=
100
R
1PA
T, make T the subject of the
formula.
6. Given that
2
bvaR +=
make v the subject of the formula
7. The formula for nding velocity is v = u +at, make time (t)
the subject of the formula.
8. In the formula,
xg
3
Tv
W =
make x the subject of the formula.
9. In the formula y = x –z
2
, make z the subject of the formula.
10. In the formula
=
c 1
x+1
make c the subject of the formula.
Glossary
192
193
A formula is an equation that shows a relationship between
two or more variables
Literal equations are equations with several variables or un-
knowns that represent a value, for example V = u+ at.
Transpose refers to “change or rearrange” in this case, the sub-
ject of the formula.
References
Chikwakwa, et al (2002), Senior Secondary Mathematics
Student′s Book 3. Blantyre: Macmillan Malawi.
Hau S and Saiti F (2010). Strides in Mathematics 3. Blantyre:
Longman Malawi.
John Fawdry (1994). Additional Mathematics a Course for
students. Mathematics Association of Malawi
Mathteacher, subject of formula, www.mathteacher.com.au/.../
trans.html retrieved on 24/06/2014
194
195
Unit
10
EXPONENTIAL AND
LOGARITHMIC EQUATIONS
In unit 6you learnt about
functions. In this unti you are
going to solve exponential and
logarithmic equations.
Exponential and logarithm
equations are used in many
elds like electrical engineering,
metrology, communication and
many elds of production.
Exponential equations
You have learnt about equations
before. Can you dene the word
equation and give some examples of
equations.
The expression such as f(x) = 2
x
is an
equation. This can also be written as
y = 2
x
. What do you call
such equation? This is known
exponential equation.
Denition
The word ‘exponential’ is an adjective
from the word exponent, which means
power or index. An exponential
equation is one which takes the form
y = x
n
, where x is the base and n is
power (an exponent).
In general:
Exponential equations are
equations in which the exponent
is the variable.
Activity1:
Expressing numbers as powers of
given base by modelling.
Find the value of each box by
completing table (b) below given on
the right hand side.
194
195
Base
0 1 2 3 4 5 0 1 2 3 4 5
2 2
0
2
1
2
2
2
3
2
4
2
5
2 1 2 4 32
3 3
0
3
1
3
2
3
3
3
4
3
5
3 81
4 4
0
4
1
4
2
4
3
4
4
4
5
4 16
5 5
0
5
1
5
2
5
3
5
4
5
5
5
To evaluate 5
3
, using calculator; follow these steps.
• Enter 5 on the calculator which is the base.
• Then punch the button with the symbol “x
y”
.
• Then punch 3 which is the power.
• You should be able to get 125.
Note Calculators differ as such you are supposed to follow its
manual. The one above is for CASIO fx- 83ES
You will now learn the reverse process.
Expressing numbers as powers of given base
You can now do the reverse process of changing a given number
to given base. First do the activity below.
Activity 2:
Expressing numbers as powers of a given base.
Express the given numbers to a given base
a. 32 to base 2
b. 27 to base 3
Share your work with your friends.
In form 1 you learned how to change a number from one base to
another. Do you remember how you were doing that?
Index
Base
Index
196
197
To change a given
number to particular
base you repeatedly
divide the number by
the base. Here is an
example.
Example1:
Change of base
Change to 32 to base 2
Divide 32 by 2
continuously and nd
how many times you
do so to get 1.
2
2
2
2
2
2
32
4
16
2
8
2
1
You have divided 32
by 2 ve times to get
1.
Hence 32 =2
5
Now do the following
exercise.
Exercise 10a
Express the following
to the given base
a. 81 to base 3
b. 243 to base 3
c. 125 to base 5
d. 16 to base 2
e. 625 to base 5
f. 1296 to base 6
g. 343 to base 7
h. 1024 to base 4
i.
to base 2
Now you look at
exponential equations.
Solving exponential
equation
You learned about
equations in the
past. Can you
write down some
equations? Recall
that an equation
is an algebraic
expression with the
left hand side equal
to the right hand
side. On the other
hand an exponential
equation is an
equation in which the
exponent (index) is
the unknown. Hence
solving exponential
equations implies
nding the value of
the exponent.
Do the following
activity.
196
197
Activity 3:
Solving exponential equation
In pairs, nd the value of x in the following:
(a) 2
x
= 8
(b) 2
x
= 32
Discuss your ndings as a class. Now see how these can be done
by going through these examples below.
Example 2:
Solving exponential equations
Solve the following exponential equations.
(a) 3
2x
= 27
Solutions:
To solve exponential equations, rstly you need to express the
left and right hand sides of the equations to the same base and
then equate the indices.
(a) 27 = 3
3
…..................... change 27 to base 3.
∴
3
2x
= 3
3
............... Equate the indices
∴
2x = 3
∴
x =
2
3
(b) 2
x
= 0.0625
0.0625 =
10000
625
…. Expressing as a proper fraction
=
4
2
1
= 2
−4
….. Simplify the fraction and change to
base 2
∴
2
x
= 2
−4
x = −4 … express to 2 and equate the bases
(c) 4
x+2
= 64
2x-1
198
199
Solving;
In this case, both 4 and 64 must be expressed to base 2
2
2(x +2)
= 2
6(2x-1)
… expressing 4 and 64 to base 2
Then 2(x+2) = 6(2x - 1) ….equate the powers
2x+4 = 12x - 6 …..Remove the brackets
12x – 2x = 6 +4 …. Arranging like terms together and
swapping
10x =10 …… subtracting and adding like terms
∴
x = 1 ……. divide both sides by 10
Now do the following exercise.
Exerise 10b:
Solving exponential equation.
1. Solve the following exponential equations
(a) 2
x
= 64
(b) 9
x
= 81
(c) 10
x
= 1000 000
(d) 5
x
= 125
(e) 2
x
= 128
(f) 3
x
=
81
1
(g) 5
x
=
25
1
(h) 4
x
= 2
(i) 64
x
= 4
(J) 10
x
= 0.001
198
199
(k) 8
x
= 64
(l) (-2) = - 8
2. Solve the following;
(a) 5
2n
– 6 × 5
n
+ 5 = 0
(b) 32
t – 1
= 192
(c) 9
p + 1
= 27
p
(d) 7
2y – 5
= 343
(e) 2
2x
– 5(2x) + 4 = 0
(f) 12 + 2
b
= 2
2b
(g) 8
2y – 2)
= 4
2y + 1
(h) 27
r – 4
= 81
2r + 2
(i) 2
2x
– 6 × 2
x
+9 = 0
In exercise 10b above, the equations are exponential equations.
Now you will solve logarithmic equations.
Logarithm equations
Let a and x be the positive numbers, a ≠ 1. The logarithm of x
with base a is denoted by log
a
x and is dened as follows; log
a
x
= y if and if a
y
= x.
The expression log
a
x is read as “log base a of x”. Thus
logarithm is another word for “power”.
We know that 10
2
= 100.
In this case “the logarithm to base 10, of 100 is 2”
How can you write this as a logarithm equation?
Activity 4:
Expressing exponential equation as logarithm equations
200
201
In pairs, express the following as logarithm equation
a. 10
2
= 100
b. 2
3
= 8
Share your work with colleague in the classroom.
Look at the following example.
Example 3:
Expressing logarithm equations
Express as logarithm equation;
a. 10
4
= 10000
Solution:
log
10
10000 = 4
This is read as; ‘the logarithm of 10000 to base 10 is 4.
b. 64 = 4
3
Solution:
log
4
64 = 3
In general;
If a
m
= P then log
a
P = m.
Now do the following exercises;
Exercise 10c
1. Rewrite as a logarithm equations.
(a) 10
3
= 1000 (b) 10
6
= 1 000 000
(c) 2
5
= 32 (d) 5
2
= 25
200
201
(e) 4
3
= 64 (f) 3
4
= 81
(g) 4
4
= 256 (h) 7
1
= 7
(i) 8
0
= 1 (j) 3
−2
=
3
4
2. Express the following from logarithm to exponential.
(a) Log
10
100 000 = 5
(b) Log
2
16 = 4
(c) Log
4
16 = 2
(d) Log
9
3 =
3
4
(e) Log
5
0.2 = − 1
(f) Log
10
10 = 1
(g) Log
3
1 = 0
(h) Log
10
0.1 = -1
(i) Log
2
6 ≈ 2.585
Activity 5:
Finding the value of a given logarithm
Discuss in pairs;
Find the value of log
2
8.
Share your work with your friends in the class room.
Now look at the examples below if it is in line with what you
have just done.
Example 4:
Finding logarithms
Find the value of
202
203
(a) log
2
64
(b) log
10
100000
(c) Log
4
0.25
(d) Log
a
(a
3
)
Solution:
Here nd the number of times the base can multiply itself to get
the number you want nd its log.
Since 2
6
= 64
∴
log
2
64 = 6. i. e multiplying 2 six times gives 64.
(b) log
10
100000
Solution
Let log
10
100000 = a
10 0000 = 10
a
changing to exponential
But 100000 = 10
5
Then 10
a
= 10
5
∴
log
10
100000 = 5
We know that 0.25 =
3
4
= 4
−1
∴
log
4
0.25 = −1
In general, if a
n
= a
m
then n = m
Do the following exercise.
Exercise 10d
Find the value of
(a) log
3
81
(b) Log
3
243
(c) log
7
343
(d) log
5
25
(e) log
12
12
(f) log
2
0.25
(g) log
10
10
(h) log
a
(a
8
)
202
203
(i) log
2
1
16
(j) log
3
4
8
1
(k) log
5
1
Rules of logarithm
When a logarithmic equation involves more than one term
containing the unknown, you need to learn how to simplify such
logarithmic expressions. To this end, you will look at the laws of
logarithms which are similar to the laws of indices.
Multiplication rule
The rst law states that ‘the logarithm of a product of two
numbers is equal to the sum of the logarithms of each of the
numbers.’
Prove that log
a
pq = log
a
p + log
a
q
Let m = log
a
p. then p = a
m
….. Changing to exponential equation
Let n = log
a
q then q = a
n
. ……changing to exponential equation
pq = a
m
×
a
n
……… substitute a
m
for p and
a
n
for q
= a
m + n
…. Law of indices, same base, powers add
Then log
a
pq = m + n = log
a
p + log
a
q
∴
log
a
pq = log
a
p + log
a
q
You should use the multiplication rule to simplify logarithm. Do
the activity 6 below.
Activity 6:
Simplifying logarithm of numbers to a given base
204
205
Express the following as single logarithms using multiplication
rule
Log
10
4 + Log
10
25
Discuss your ndings as a class.
Now look at the following examples;
Example 5:
Express the following as single logarithms
(a) log
3
6 + log
3
7
Solution
log
3
(6
×
7) Using the multiplication law
= log
3
42.
(b) log
6
6 +log
6
2+ log
6
3
Solution;
log
6
(6 × 2 × 3) = log
6
36
Using multiplication law.
2.
Division rule
The second law states that ‘the logarithm of a quotient is
equal to the logarithm of the dividend minus the logarithm
of the divisor’
Prove that log
a
3
4
= log
a
p − log
a
q.
Let m = log
a
p. Then p = a
m
….. Changing to exponential
equation
Let n = log
a
q then q = a
n
…… changing to exponential
equation
3
4
=
n
m
a
a
……. Substitute a
m
for p and a
n
for q
= a
m – n
… division rule of indices
204
205
Then log
a
3
4
= m − n = log
a
p − log
a
q. …. substitution
∴
log
a
3
4
= log
a
p − log
a
q.
Look at the given examples below.
Example 6:
Expressing as single Logarithms
Express the following as single logarithms
(a) log
2
15 − log
2
5
Solution:
log
2
(15
÷
5) Using the division law
= log
2
3
(b) log
10
175 – log
10
25
Solution:
log
10
(175
÷
25)
= log
10
7
Powers
Prove that log
a
p
n
= n log
a
P.
Let m = log
a
p . Then p = a
m
….. Changing to exponential
equation
P
n
= (a
m
)
n
….............................. Since p is raised to n
= a
mn
….................................. law of indices
Then log
a
P
n
= mn = n log
a
P. …... Substitute
∴
lo
a
p
n
= n log
a
P.
206
207
Example 7
Express the following as single logarithms
(a) 2 log
5
3 + 3 log
5
2
Solution
2 log
10
3 = log
10
2
= log
5
9
3 log
5
2 = log
5
2
3
= log
5
8
∴
2 log
5
3 + 3 log
5
= log
5
9 + log
5
8
= log
5
(9
×
8) Using the multiplication law
= log
5
72
(b) Log
10
3 − 2 log
10
( )
Solution
Log
10
3 − 2 log
10
( )
2
..................log rule
= log
10
3 − log
10
16
1
= log
10
(3
÷
16
1
) ......Using the division law.
= log
10
48.
The base and number the same
Prove that log
a
a = 1
Let log
a
a = n then a = a
n
to exponential equation
∴
a
1
= a
n
n = 1 since the bases are the same
∴
log
a
a = 1
Example 8:
1
4
1
4
4
3
206
207
Express as single logarithms
log
10
150 - log
10
15
Solution
log
10
(15
÷
15)
log
10
10 = 1
Note. The logarithm of numbers of the same base is always 1.
The log of 1
Prove that log
a
1 = 0
Let log
a
1 = m
1 = a
m
But a
0
= 1.
∴
a
m
= a
0
∴
m = o
Hence log
a
1= 0.
Note The logarithm of 1 is always 0.
Exercise 10e
. Write as a single logarithm
(a) log
3
6 + log
3
4
(b) log
2
48 − log
2
6
(c) 3log
5
2 + log
5
10
(d) 2log
6
8 − 4log
6
3
(e) Log
10
5 + log
10
6 − log
10
(f) log
2
0 – log
2
(g) log
2
14 – log
2
7
(h) 2log
a
3 + 3log
a
2 –log4
(i) log
3
8.1 + log
3
10
(i)
4
1
1
8
log 25
log5
Example 8
208
209
Sometimes, you may require to perform the reverse process.
Example 9:
Expressing logarithms
1. Write the following in terms of log
a
x, log
a
y, and log
a
z
(a) log
a
x
3
y
2
z
4
(b) log
a
2
3
y
zx
Solution
(a) log
a
x
3
y
2
z
4
= log
a
x
3
+ log
a
y
2
+ log
a
z
4
using the
multiplication law.
= 3log
a
x +2log
a
y + 4log
a
z using the power law.
(b) log
a
2
3
y
zx
= log
a
x
3
+ log
a
8
− log
a
y
2
using the
multiplication and
Division laws
= 3log
a
x +
3
4
log
a
z − 2log
a
y since √z = z
1/2
using the power
law.
2. Given that log
10
2 = 0.431 and log
10
3 = 0.683, nd the value
log 18
Solution:
Log
10
18 = log
10
2
3
+ log
10
3
= 3log
10
2 + log
10
3
= 3(0.431) + 0. 683
=1.976
208
209
Exercise 10f
1. Write in terms of log
a
x, log
a
y and log
a
z
(a) log
a
()
53
zxy
(b) log
a
−4x
x
(c) Log
3
y
zx
(d) log
a
( )
33
×a
2. Given that log
5
2 = 0.431 and log
5
3 = 0.683, nd the value of
(a) log
5
6 (b) log
5
1.5
(c) log
5
8 (d) log
5
12
(e) log
5
18
1
Having studied the laws of logarithms, you can now solve
logarithmic equations with the unknown in more than one term.
Activity 7:
Solving log equation
(a) In pairs, solve the following
(i) x = log
2
8
(ii)log
2
x = −3
(iii)log
x
25 = 2
Present your work to others members in the classroom.
(e) log
a
z
yx
(f) log
a
−4x
x
(g) log
a
5
a
x
210
211
In this activity, you change from logarithm equation to
exponential equation and then solve.
Example 10:
Solving logarithmic equations
Solve the equations below
(a) log
5
x = 1 + log (x − 4)
(b) 2log
3
x = log
3
(x + 6)
(c) log
2
x = −3
Solution
(a) Since there is one term on each side, we can take the
antilogarithm on both sides
∴
antilog (log
3
x) = antilog (log
3
(2x − 6).
∴
x = 2x – 6. Arrange like terms together
∴
x = 6
(b) Collect the logarithm terms to one side
i.e. log
5
x − log
5
(x − 4) = 1
∴
log
5
−4x
x
= 1
∴
log
5
−4x
x
= log
5
5 (since log
5
5 = 1)
then take antilogarithm on both sides
∴
a
b
2
−
= 5
∴
x = 5 (x − 4) … multiply (x- 4) both sides
∴
x = 5x − 20
∴
4x = 20 …… like terms together and divide both sides by
∴
x = 5
210
211
(c) log
3
x
2
= log
3
(x + 6)
∴
x
2
= x + 6 …….take antilog both sides
∴
x
2
− x − 6 = 0…. arrange like terms together
∴
(x − 3) (x + 2) = 0…. Factorise LHS and solve for x
∴
x = 3 x = − 2
∴
x = 3 as it is not possible to take log (−2)
Exercise 10g
Solve the following logarithmic equations.
(a) x = log
3
27 (b) x = log
5
625
(c) log
x
125 = 3 (d) log
x
81 = 2
(e) x = log
2
8
1
(f) x = log
169
13
(g) log
3
1 = x (h) log
2x
36 = 2
(i) log
4
(x − 2) = 3
(j) log
x − 1
8 = 3.
(k) log
2
256 = x
Real life problems of exponential and logarithm equations
As earlier pointed out, exponential equations are used for even
more contexts, including population and bacterial growth,
radioactive decay, compound interest, cooling of objects, and
growth of phenomena such as virus infections, Internet usage
and many others.
Example 11
Populations
a. A tree frog population doubles every three weeks. Suppose
that currently, there are 10 tree frogs in your back yard.
How many tree frogs will there be in 6 months, assuming
that there are four weeks each month?
212
213
b. How long will it take this population to be 10,240?
Solutions
a. First gure out how many times this population will double
in 4 months. Each month 4 weeks, then six months 6x4= 24
weeks.
Since the population doubles every three weeks, then
24/3 =8 times in 24 weeks. Look at the table below:
Number of
weeks
Doubling
period
population
O 0 10
3 1 20 =10 x 2or 10 x 2
1
6 2 40 = 10 x 2 x 2 or 10 x 2
2
9 3 80 = 10 x 2 x 2 x2 or 10 x 2
3
After 24 weeks, the population will be 10 x 2
8
= 2560 tree
frogs!
If you look at the table above, you will notice that you
could let n be the number of weeks. How would you get the
number of doubling periods from n?
Solution:
After n weeks, the population would be P = 10 x 2
(n/3)
.
So you need to solve the equation
10,240 = 10 x 2
(n/3)
.
1024 = 2
(n/3)
… divide both sides by 10,
But 1024 = 2
10
….expressing to base
2
10
= 2
(n/3)
. … equate the bases.
10 = n/3,
∴
n = 30.
That means that after 30 weeks, the population will be
10,240.
212
213
Unit summary
• In this unit so far, you have looked at exponential and
logarithm equations. You have learned how to express
a number as a power of a given base. You also modelled
exponential equations and solved them. Furthermore, you
looked at rules of logarithm and also solved logarithm
equations. In the next unit you will learn about triangles in
trigonometry.
Unit review exercise
1. Solve the following equations.
(a) 2 log
7
x = log
7
(x + 2)
(b) log
5
x = 1− log
5
(x − 4)
(c) log
3
(x + 3) = 2 log
3
(x + 1)
(d) log
7
2x = log
7
(x + 2)
(e) log
4
2 − log
4
x = log
4
2/3.
(f) log
2
(x + 1) = 1
(g) log
3
6 = log
3
3 +log
3
x
(h) 8
x-1
= 16
(i) 3
2y
- 4(3
y
) + 3 = 0
(j) 2
2x
– 4(2
x
) + 3 = 0
(k) 25
2b
÷5
b
= 5
6
2. Currently, 80,000 bacteria are present in a culture. When
an antibiotic is added to the culture, the number of bacteria
is reduced by half every 3 hours.
(a) How many bacteria are left after a day?
(b) When will fewer than 1000 bacteria be present?
3 write as a single number and simplify if possible;
(a)
log16+log4
log16
log4
(b)
log 3+log27
log9
log3
214
215
Glossary
Base is a number that is being raised to a power
Exponent is a number that has been raised as a power known
also as power or indice
Logarithm equation is an expression of the form log
a
x = y
which is read as “log base a of x equals y”
Exponential equations is an equation in which the exponent
is the variable.
References
Hau S and Saiti F (2010). Strides in Mathematics 3. Blantyre:
Longman Malawi.
Elaine Ryder, Paul McAdams, Pat Huddleston, (2013), CHANCO
Teach yourself series, Mathematics, Second Edition, Chancellor
College Publication, Zomba.
Hardwood Clarke and Norton F G J, (1984) Seventh Edition,
Heinemann Educational Book Ltd, Oxford, London
214
215
Unit
11
TRIGONOMETRY
In your JCE course you learnt
about calculating sides of a right
angled triangle using Pythagoras
theorem. In this unit you will
learn how to calculate sides and
angles of a right angled triangle
using trigonometry. You will also
learn how to derive trigonometric
ratios of 30
0
, 45
0
, 60
0
, 90
0 Z
and
solve practical problems involving
trigonometry.
The knowledge gained in learning
this will be used in solving real
life situations such as nding the
height of a mountain.
Activity 1:
Identifying the sides of a right
angled triangle
You are familiar with the side of the
triangle called the hypotenuse. You
will now be introduced to the names
of two other sides in relation to the
angles of the triangle.
In groups, study the denitions
below which relate to a right angled
triangle:
• An adjacent side to an angle is
the side which together with
the hypotenuse forms that
angle.
• An opposite side to an angle
is a side which together with
the adjacent side forms a right
angle of the triangle.
Basing on these two denitions,
identify the adjacent side and the
opposite side to angle A in the
drawing below:
A
B
C
Which side is the hypotenuse in the
above triangle?
Come up with your own right angled
triangles labeling them with letters
of your choices and identify the
hypotenuse, the opposite sides, and
the adjacent sides in relation to the
angles of the triangles. Let your
teacher check your answers.
216
217
Trigonometric ratios
In activity 1 you learnt to identify the sides of a triangle in
relation to the given angles. Any two of these sides can be
divided and the result is called the trigonometric ratio. You will
now learn to dene three of these ratios, sine ratio, cosine ratio
and tangent ratio.
Activity 2:
Deriving sine ratio
In your groups,
1. Draw the following right angled triangles: 4cm by 3 cm by
5cm; 8cm by 6cm by 10cm; 12cm by 9cm by 15cm.
2. Label the angle opposite the shortest side in each triangle
as θ (theta).
3. Now in relation to angle θ, nd the ratio of the opposite side
to the hypotenuse by dividing the side opposite angle θ by
the hypotenuse in each of the three triangles simplifying
the fractions to their lowest terms (or give the answer as
decimal fractions to 4 decimal places). What do you nd?
4. Report your ndings.
5. Draw your own right angled triangles and practice getting
sine ratio and in your group draw one triangle on a chart
paper and hang the chart on the wall of your classroom.
The ratio you found above is called the sine ratio and as you
have seen, it doesn’t depend on the size of the triangle. From
this, can you try to dene what the sine of an angle is?
Because you found this ratio using angle θ you specically call it
sine of angle θ or in short Sin θ.
Activity 3:
Deriving cosine ratio
1. Go back to the triangles you drew in activity 2. Repeat
216
217
step 3 but now divide the side adjacent angle θ by the
hypotenuse. Again simplify the fraction to their lowest
terms.
2. Report your ndings.
The ratio you found is called the Cosine ratio and because you
found this ratio using angle θ you specically call it Cosine of
angle θ or in short Cos θ. Again, it doesn’t depend on the size of
the triangle. As you did in activity 2, can you try to dene the
cosine of an angle?
Activity 4:
Deriving tangent ratio
1. Again go back to the triangles you drew in activity 2. Repeat
step 3 but now divide the side opposite angle θ by side
adjacent angle ө simplifying the fraction to its lowest term.
2. Report your ndings.
The ratio you found is called the tangent ratio. As in sine and
cosine ratios, it doesn’t depend on the size of the triangle.
The three trigonometric ratios above can easily be remembered
by the word SOHCAHTOA. Discuss with a friend to see how this
word has been formed. Ask your teacher to help you if you have
problems.
Exercise 11a
1. In each of the following triangles nd the sine of angle A
(Sin A): Give your answers correct to 4 decimal places:
218
219
C A
7.5m
10.9m
B
A
12.1 cm
14.1cm
B 7.3cm C
(a) (b)
2. In each of the following triangles nd the cosine of angle
(Cos X): Give your answers correct to 4 decimal places.
X
12cm 6cm
Y Z
(a)
Z
8.7cm
X 3.4cm Y
(b)
3. Find the tangent of angle PQR in each of the following
diagrams. Give your answers correct to 4 decimal places.
218
219
Q 8cm P
R
13cm
P
9cm
R
7cm
Q
P 5cm R
4cm
Q
(a)
(b)
(c)
4. Given that in a right angled triangle ABC, the cosine of
angle A = 0.8, angle B = 90
0
nd
(a) Sine of angle A
(b) Cosine of angle C
(c) Tangent of angle C
5. Draw your own triangles and write down sine, cosine and
tangent of angles of your choices. Get a friend or your
teacher to check your work.
Using a scientic calculator to work out problems involving
trigonometry
You will quite often use a calculator to work out problems
involving trigonometry in the sections that follow. You will use
it especially in nding the tangent, sine and cosine of angles and
nding angles whose ratios are given. The aim of this section
therefore is to help you acquire the skills in using the calculator
in these areas.
220
221
Activity 5:
Using a scientic calculator to nd tangent, sine and
cosine of angles
Before starting to work with your calculator, ensure that the
calculator is in degree mode. Ask your teacher to help you set
your calculator in this mode.
In pairs,
1. Study the keys on your scientic calculator. How are sine,
cosine and tangent represented on the calculator?
2. Now press the “on” key and press “Cos” key followed by 53.
What does your calculator display?
3. Now press the “=” key and write down the display. You can
correct this display to four decimal places. This result is the
cosine of 53
0
.
4. Now using your calculator, nd the following: tan 60
0
, tan
40
0
, sin 30
0
, sin 90
0
.
5. Choose your own angles and nd their sines, cosines and
tangents using the calculator.
Exercise 11b
Use a scientic calculator to nd, correct to four decimal places
1. Tan 30
0
2. Cos 47
0
3. Sin 35
0
4. Tan 80
0
5. Sin 75
0
6. Cos 76
0
7. Sin 33
0
220
221
8. Tan 24
0
9. Cos 23
0
10. Sin 48
0
Activity 6:
Using a scientic calculator to nd angles whose
trigonometric ratios are given
Again in pairs, identify the key on the scientic calculator
labeled “2
nd
” or standing against “shift”. You are to use this key
together with other keys to nd angles whose trigonometric
ratios are given.
1. Press the key identied above and then press Cos. Write
down what your calculator displays.
2. Repeat this step but now use Tan and Sin instead of Cos.
Again write down what the calculator displays.
3. Discuss what the displays mean. Ask your teacher to help
you if you cannot come up with the meaning.
4. In activity 5, you found Cos 53
0
as 0.6018 to four decimal
places. Now press the 2
nd
function key on your calculator
followed by Cos followed by 0.6018 followed by “ = ” key.
Write down what the calculator displays to the nearest
degree.
5. Discuss the order of pressing the keys if you were to nd the
angle whose tangent or sine is given.
6. Repeat step 4 but now using the ratios you found in step 4
of activity 4. Did you manage to get the angles you started
with?
7. In your own words write down the statement for nding
angles whose trigonometric ratios are given.
Exercise 11c
Use a scientic calculator to nd to the nearest degree the angle
222
223
whose
1. Cosine is 0.5411
2. Cosine is 0.0089
3. Tangent is 2.3412
4. Sine is 0.7698
5. Tangent is 1.2300
6. Sine is 0.1543
7. Tangent is 0.4567
8. cosine is 0.7899
9. Sine is 0.9765
10. Tangent is 0.8354
Calculating sides and angles of a right angled triangle using
trigonometry
In Form 2, you learnt how to calculate sides of a right angled
triangle using Pythagoras Theorem. Remember that to use
Pythagoras theorem you must be given two sides. However you
may sometimes be given an angle or angles and a side and you
may still be required to calculate the sides of a right angled
triangle. This is when you can use trigonometry. In the activities
that follow you will learn how to calculate sides and angles of a
right angled triangle using trigonometry.
Activity 7:
Calculating sides of right angled triangles using
trigonometric ratios
You learnt how to draw triangles in your JCE course using a
ruler, a protractor and a pair of compasses. In groups,
1. Construct triangle PQR in which QR = 4cm, angle R =
90
0
and angle P = 32
0
.Each one must draw his or her own
222
223
triangle.
2. Measure and record the lengths of PQ and AR.
3. Compare your work.
4. Report your ndings.
You could also use scientic calculators to work out the lengths
of PQ and AR. Again in your pairs,
1. Draw triangle PQR again. Just draw the triangle without
using a compass or a protractor.
2. Using the triangle write down the relationship between 32
0
and the two sides you are to nd.
3. Using the words hypotenuse, adjacent and opposite ,
complete the following:
Tan 32
0
=
Cos 32
0
=
4. Discuss how you could solve the two equations to get the
unknown sides using a calculator.
5. Using the calculator, nd the unknown lengths. Compare
your answers to the answers you obtained in step 2 in the
triangle you constructed. Comment on your results.
To nd the lengths of a triangle using trigonometry you need
to rst see the relationship between the given angle(s) and the
given side and the side you are to nd. This helps you to come
up with the more direct trigonometric ratio to use. For example
if the relationship is adjacent and hypotenuse, then you use
Cosine ratio because it is the one which uses these sides. You
then proceed as in the s that follows:
Example 1:
Calculating sides of a triangle
224
225
A
C
2.4 cm
B
74
0
In the above triangle, nd the length of AB.
Solution:
The relationship between 74
0
and 2.4cm and AB is adjacent and
hypotenuse so you use Cosine ratio as follows:
Cos 74
0
=
2.4cm
AB
--------- from the denition of Cosine of an angle.
AB Cos 74
0
= 2.4 --------- multiply both sides by AB
AB =
2.4cm
cos 74
0
------------------------ divide by Cos 74
0
both sides
AB = 8.71 cm approximately
Challenge
Discuss other ways in which the above question could have been
solved.
Example 2
W
X
33
0
9.8cm
Y
224
225
Calculate the length of WX in the above gure correct your
answer to one decimal place.
Solution:
The relationship between 33
0
and 9.8cm and WY is opposite
and hypotenuse so you use sine ratio as follows:
Sin 33
0
=
WY
9.8cm
------------From the denition of sine of an angle
WY = 9.8 Sin 33
0
cm------Multiply by 9.8cm both sides and
rearrange.
WY = 5.3 cm.
Example 3
A straight pole is rested against a wall with its one end 2.3m
from the foot of the pole. The other end rests on top of the wall
and the pole makes an angle of 65
0
with the wall. Calculate the
length of the pole to the nearest metre.
Solution
You rst need to make a sketch of the information. Straight
lines are used to represent the wall and the ground on which the
wall is standing. The assumption is that the wall and the ground
are perpendicular.
65
0
Wall Pole
2.3m
Sin 65
0
=
2.3m
Lenght of thepole
------------------------ From the denition of
sine of an angle
226
227
Length of the pole(Sin 65
0
) = 2.3 m--------- Multiply by length of
the pole both sides
Length of the pole =
2.3
65
0
---------------------------- Divide by Sin 65
0
both sides
Length of the pole = 3m (to the nearest metre) ----- From the
calculator.
Exercise 11d
1. Calculate, correcting your answers to 1 decimal place, the
unknown sides in each of the triangles below:
(a)
R
S
T
=
=
=
7.5cm
35°
(b)
P
11.2cm
R
Q
=
49°
2 A straight pole 10.5m long is leaning against a
perpendicular wall. If the pole makes an angle of 65
0
with
the top of the wall ,calculate the height of the wall.
3 A rectangle is 10m wide. One of the diagonals of the
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rectangle makes an angle of 36
0
with the shorter side of the
rectangle. Calculate the length of the rectangle.
4. Two wires are tied to the top of a pole which is standing
perpendicular to the ground. They are then straight xed
to the ground along the same line so that the shorter
wire makes an angle of 65
0
with the ground and the angle
between the two wires is 10
0
. Given that the shorter wire is
8.5m long, calculate the distance between the two wires on
the ground.
5. A rectangular room is 10.2m long. The diagonals of the oor
of the room each make an angle of 22
0
with the length of
the room.
Calculate
(a) the length of the width of the room.
(b) if the diagonal from the top of the corner to the
bottom opposite corner of the room makes an angle of
60
0
with the height of the room, calculate the height
of the room.
Activity 8:
Calculating angles of a right angled triangle using
trigonometry
In the previous activities, you found sides of a right angled
triangle by measurement and by using a calculator.
In this activity you will learn how to nd angles of a right angled
triangle using a calculator.
1. Draw triangle XYZ in which XY = 3cm, YZ = 4.8 cm, XZ =
5.6 cm and angle Y = 90
0
2. Write down the ratios for sin X and tan Z.
3. Discuss how you could use a calculator to nd the values of
angle X and angle Z. If you nd problems go back to activity
5 of this topic.
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Example 4:
Calculating angles
Y 5.2cm X
4.5 cm
Z
Find the value of angle XYZ in the triangle.
Solution:
There are more than one ways of solving this problem. Here is
one of them:
Tan angle XYZ =
4.5cm
5.2cm
Tan XYZ = 0.8653 (correct to 4 decimal places)
Angle XYZ= tan
-1
0.8653
Angle XYZ = 30
0
approximately (from a calculator)
Challenge: Find other methods of solving this problem and let
your teacher check the methods.
Exercise 11e
13.5cm
19cm
1.
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Calculate to the nearest degree, the value of angle marked θ
(theta) in the above triangle.
2. In a right angled triangle PQR, PQ = 6 cm, PR = 15 cm
and angle PQR = 90
0
. Without using Pythagoras theorem,
calculate the other two angles of the triangle giving your
answers correct to the nearest degree.
3. A pole is resting in the corner of the room 4m by 5m. The
top of the pole is 6.8m above the oor and the bottom is
2.5m from each wall. Calculate the angle that the pole
makes with the oor.
4. A wheelchair ramp is to be built over steps up to a college
entrance. Each step has a vertical rise of 12cm and a
horizontal tread of 45cm. Calculate , to 1 decimal place, the
angle that the ramp makes with the horizontal.
5. Madalitso tries to row straight across a river which is 46m
wide. The current carries her downstream at an angle of 72
0
to the bank. How far downstream from the point she was
trying to reach does she actually land?
Trigonometric ratios of special angles
Special angles are angles whose trigonometric ratios can be
expressed as surds or as simple fractions. These angles are
30
0
,45
0
,60
0
and 90
0
. The trigonometric ratios of these angles help
us to solve problems involving trigonometry without having to
use a calculator. In the activities that follow, you will learn how
to derive the trigonometric ratios of these angles.
Activity 9:
Deriving fractional trigonometric ratios of 30
0
and 60
0
In groups,
1. Draw and label any one equilateral triangle.
2. From any one vertex in each triangle draw a perpendicular
bisector of the side opposite that vertex.
3. What is the value of each angle in each triangle?
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4. Work out the following ratios: sin 30
0
,cos 30
0
, sin60
0
and
cos60
0
. Leave your answers as simplied surds or as
simplied fractions.
5. Summarise the information in the table below by lling the
trigonometric ratios you found in step 4 above:
Sin 30
0
Cos 30
0
Tan 30
0
Sin 60
0
Cos 60
0
Tan 60
0
You must have discovered that despite the groups drawing their
own equilateral triangles, all the trigonometric ratios reduce to
the same results. However the amount of working varies from
one triangle to another. If you draw an equilateral triangle
whose length is an odd number you would work more than it
was if the length was an even number. Furthermore, if you want
to reduce the amount of working even further, it is advisable to
use the smallest equilateral triangle of even numbered length of
2 units.
Activity 10:
Deriving trigonometric ratios for 45
0
In groups, draw any three isosceles right angled triangles.
1. Use Pythagoras theorem to calculate the lengths of the
hypotenuse in each of the triangles, leaving them in
simplied surd form.
2. Work out the following ratios for all the three triangles:
tan 45
0
, cos 45
0
and sin 45
0
. Simplify your answers to their
simplest forms.
3. Summarise the information in the table below:
Tan 45
0
Cos 45
0
Sin 45
0
As in activity 9, you must have discovered that for any right
angled isosceles triangle the fractional trigonometric ratios are
the same.
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Activity 11:
Deriving trigonometric ratios for 90
0
1. Working in groups, draw the y and x axes on a square sheet
of Cardboard of not less than 40cm by 40cm.
2. Get two straight sticks each 30cm long, and pivot the two
ends of the sticks.
Ensure that the two sticks can move freely on the pivot. Bring
the two sticks onto the cardboard and pivot one end of the two
sticks at the origin as shown below:
y-axis
x-axis
pivot centre
Square cardboard
O
S
R
30cm
30cm
T
3. Now pin the cardboard on to a wall so that TR is
perpendicular to the x axis. Note that OS is the
perpendicular distance of TR from the y axis.
4. Now write down the following ratios of acute angle O in
terms of OT and OS: Cos angle TOS, Sin angle TOS and
Tan angle TOS.
5. Slowly, rotate OT anticlockwise about O. What is happening
to the size of acute angle O? What about the distance OS?
6. Continue rotating OT anticlockwise until R , S and O
coincide. What will be the size of the acute angle O at this
moment? What will be the length of OS?
7. Using your ndings in step 6, make appropriate
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substitutions into the ratios you found in step 4, simplify
the results and give a summary in a table form.
Challenge
In deriving trigonometric ratios for 90
0
, you used sticks of length
30cm. Do you think you would obtain different results if sticks
of other lengths say 10cm and 10cm, 20cm and 20cm e.t.c were
used?
The trigonometric ratios of special angles are useful in solving
some right angled triangles without having to use a calculator.
Example 5:
Solving for triangles
cm
3cm
SM
M
=
x
Find the value of x without using a calculator or tables.
Solution
Cos 60
0
=
3cm
x
x cos 60
0
= 3cm ---------- multiply both sides by x
x =
3cm
Cos60
0
------ divide both sides by cos 60
0
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x =
3cm
1
2
-------- substitute cos 60
0
for (activity 9)
x = 6cm
Exercise 11f
1.
P
Q
S
R
45
0
60
0
In the triangle above, PS is the height of the triangle PQR and
QS = 40 cm.Without using a calculator, calculate, leaving your
answer in surd form, the length of PR.
2. Calculate the lengths of the labeled sides in the triangles
below leaving your answers in surd form:
(a)
5cm
x
30
0
(b)
x
10cm
60
0
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(c)
y
3
2
cm
45
0
3.
Q
8cm
60
0
45
0
T
S
R
In the above gure, calculate, leaving your answers in surd
forms, the lengths of TQ, RQ and RS.
Angles of elevation and depression
An angle of elevation is the angle between a horizontal line and
a straight line drawn above the horizontal line.
An angle of depression is the angle between a horizontal line
and a straight line drawn below the horizontal line.
The two angles are shown below:
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Angle of depression
Angle of elevation
Calculating angles of elevation and depression
Basically what happens in calculating angles of elevation and
depression is what you did in activity 8 of this topic.
Example 6
Calculate the angle of elevation to the top of a building 10 m
high from a point 15m away from the foot of the building.
Solution
You need a sketch of the information. The building is assumed
to be perpendicular to the ground.
Top (T)
10m
Foot (F) 15m
Point (P)
You are required to calculate angle TPF.
Tan angle TPF =
10m
15m
Tan angle TPF = 0.6667
Angle TPF = 34
0
( to the nearest degree)
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Example 7:
Angle of depression
Calculate the angle of depression of a point 25.6 m away from
the top of a tower 15.8m high.
Solution
Horizontal (H)
Top (T)
25.6m 15.8m
Point (P) Foot (F) of a towe
You are required to nd angle HTP.
Angle HTP = angle TPF
Sine angle TPF =
15.8m
25.6m
Sine angle TPF =0.6172
Angle TPF = 38
0
The angle of depression is therefore 38
0
Exercise 11g
Give your answers correct to the nearest degree in all the
questions
1. Calculate the angle of elevation of the top of a mountain
300m high from a point 1.5Km from the foot of the
mountain.
2. Find the angle of elevation of the tip of an aerial 1.5m high
standing on a building 5m high from the point X 300m away
from the foot of the building.
3. A point is 150m away from the foot of a tower. It is also
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measured that this point is 350m from the top of the tower.
Calculate the angle of depression of this point from the top
of the tower.
4. A boy in a fruit tree 5.8m high throws a fruit straight to
his friend who is standing 3m from the foot of the tree. If
the boy on the ground catches the fruit at a height of 0.5m
from the ground, at what angle to the tree did the other boy
throw the fruit?
5. Two wires are tied to the top of a pole 10.5m high and
then pegged straight on the ground. The two wires reach
distances of 1.5m and 2m on the ground respectively.
Calculate the angle between the two wires.
6. In a penalty shootout, a player shoots a ball straight onto
a goalpost whose crossbar is 1.5 m above the ground. If the
penalty spot is 11m away from the goal line, what is the
maximum angle from which the player can score?
7. A plane ying at an attitude of 1500m is to land at an
airport 10km away. At what minimum angle must the pilot
lower his plane?
8. In a shoot a target competition, a shooter stands 35m away
from the foot of a pole on which a target is placed. If a gun
is held by the shooter at a height of 0.75m above the ground
at an angle of 20
0
and if the pole is 12m high, show that the
shooter will miss the target.
9. A rubber bullet is red at an angle straight on to a
perpendicular wall and hits a point 5.6m high. If the bullet
was red from the foot level of the wall and rebounds at an
angle of 80
0
to its path onto the wall and at 30
0
to the wall,
at what angle and distance from the wall was the bullet
elevated?
10. The tip of the roof of a room stands 1.5 m midway above
the last line of the room 6m wide. Calculate the angle of
elevation of the roof.
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Further problems involving angles of elevation and
depression
Example 8:
Problems involving angles of elevation and depression
A motorist travels a distance of 300m up a hill inclined at an
angle of 10
0
to the horizontal. If the motorist was originally at
sea level, nd the height she is above sea level at the end of
300m.
Solution
It is important to make a sketch of the information. For
purposes of calculations we use straight lines to represent
objects.
10°
B
C
300m
A
sin 10
0
=
300
300sin 10
0
= BC --------------------------------- (multiply by 300
both sides).
BC = 52.1m approximately.
Challenge
Discuss some other ways of solving this problem.
Example 9
The base of a tree is 60m away from point x on the ground. If the
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angle of elevation of the top of the tree from x is 40
0
. Calculate
the height of the tree. Give your answer to the nearest metre.
40
0
60m
X
Let the height of the tree be h
∴
cm
h
60
= tan 40
0
h = 60m
×
tan 40
0
= 50 .3m (to 3 signicant gures)
Example 10:
Problems involving angles of elevation and deviation
A person on top of a cliff 50m high, observes the angle of
depression of a boy to be 30
0
. If he is in line with the boy,
calculate the distance between the boy and the foot of the cliff
(which may be assumed to be vertical).
Horizontal
50m
60
0
30
0
X cm
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Let the distance between the boy and the foot of the cliff be x.
The angle between horizontal line and vertical cliff is 90
0
∴
to nd the angle for calculating x in the triangle, we subtract
30
0
from 90
0
i.e. 90
0
− 30
0
= 60
0
∴
50
x
= tan 60
0
x = 50
×
tan 60
0
x = 86.6 m (to 3 signicant gures)
Exercise 11h
1. From a point, the angle of elevation of a tower is 30
0
. If the
tower is 25m distant from the point, what is the height of
the tower?
2. A woman 1.6m tall observes the angle of elevation of a tree
to be 26
0
. If she is standing 20m from the tree, nd the
height of the tree.
3. A boy 1.2m tall is 10m away from a tree 20m high. What is
the angle of elevation of the top from his eyes?
4. A and B are two villages. If the horizontal distance between
them is 12km and the vertical distance between them is
2km.
Calculate
(i) the shortest distance between the two villages
(ii) The angle of elevation of B from A.
5. A surveyor stands 100m from the base of a tower on which
an aerial stands. He measures the angles of elevation to the
top and bottom of the aerial as 58
0
and 56
0
. Find the height
of the aerial.
6. A man 1.2m tall standing on top of the mountain 1200m
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high observes the angle of depression of a steeple is 43
0
.
How far is the steeple from the mountain?
7. X and Y are two towns. If the vertical distance between
them is 10km and the angle of depression of Y from X is 7
0
,
Calculate:
(i) the shortest distance between the two towns
(ii) the horizontal between the two towns.
8. An air plane receives a signal from a point X on the ground.
If the angle of depression of point X from the airplane is 30
0
,
calculate the height at which the plane is ying given that
the plane is 6km from X.
9. A girl sitting on a hill at A, overlooking a lake can see a
small boat at a point B on the lake. If the girl is at height of
50m above B at a horizontal distance of 120m away from B,
calculate:
(i) The angle of depression of the boat from the girl
(ii) The shortest distance between the girl and the boat.
10. A plane is ying at an altitude of 8km directly over the line
AB. It spots two boats A and B, on the sea. If the angles
of depression are 60
0
and 30
0
respectively, calculate the
horizontal distance between A and B (two possible answers).
11. Mr Phiri uses 10m planks to ofoad some items from his
lorry. If the lorry is 1.5m high, calculate the minimum angle
of inclination of the planks.
Bearing
Activity 12:
Using a compass to name directions
In form 1geography, you learnt about how to use a compass.
1. Name the four cardinal points on a compass.
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2. What is the initial reference line on the compass?
3. Discuss how you name points midway between the cardinal
points. Are these the only points you can show on a
compass?
This method of naming points is used to show direction.
Sometimes the cardinal points will be used together with angles.
Bearing quoted in this way are always measured from N and S
and never from E and W.
Example11:
Naming directions
State the directions of the huts in each of the following diagrams
(a)
N
W
E
S
60
0
Solution:
N60
0
E
(b)
N
W
E
S
71
0
Solution
S71
0
W
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Bearings are also measured from north in a clockwise direction,
the north being taken 0
0
. Three gures are always stated. For
example 008
0
is written instead of 8
0
. East will be 090
0
, South
180
0
and West 270
0
.
Activity 13:
Calculating the bearing of a point relative to a given
point
1. On a piece of paper draw a north-south line.
2. Draw any object on the same piece of paper and join the
object to the north - south line by a straight line. Let the two
lines join at a point say O.
3. Use your compass to nd the bearing of the object from O.
4. Your teacher will provide you with diagrams showing
locations of point a point B in relation to another point A.
5. Use your compass to nd the bearing of B from A.
Challenge
In the drawings provided by your teacher, how would you nd
the bearing of A from B?
Exercise 11i
1. Write each of the following as three-gure bearings
a) N50
0
E
b) N50
0
W
c) S50
0
W
d) S50
0
E
e) S80
0
E
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2. Find the three-gure bearing of Q from P in the sketches
below.
a)
N
Q
P
80
0
b)
60
0
Q
P
N
c)
Q
N
P
70
0
d)
Q
N
P
45
0
3. Draw sketches representing the following:-
A is on a bearing of
a) 020
0
from B
b) 125
0
from B
c) 220
0
from B
d) 270
0
from B
e) 310
0
from B
Calculating the bearing of a point relative to a given point
using trigonometry
Problems involving bearing may be solved by making a scale
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drawing or by using trigonometry. When using the scale
drawing method, you rst choose a suitable scale. Since we have
just studied trigonometry, we will solve the problems below
using the trigonometry method.
Example 12
A boat starts from point A and sails to point B on a bearing of
030
0
, given that the distance of A to B is 80m, nd how far B is
to the east of A.
Solution
You need to sketch the information. Form a right-angled
triangle so that you can use SOHCAHTOA.
30
0
80m
A
C
B
60
0
We are asked to nd AC
m
AC
80
= Cos60
0
AC = 80m
×
Cos60
0
AC = 40m
Example 13
A ship sets out from a point A and sails due north to a point B,
a distance of 120km. It then sails due east to a point C. If the
bearing of C from A is 037
0
,
Find
a) the distance BC
b) The distance AC.
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247
Solution
Again, you need to sketch what is going on. Join A to C to form
the right-angled triangle ABC.
C
A
B
N
120km
37
0
a)
120
BC
=tan37
0
bc = 120km
×
tan37
0
= 90.4km. (to 3 signicant gures)
b) To nd AC you can use the Pythagoras theorem or
trigonometry. Using trigonometry,
AC
km120
= cos37
0
∴
AC =
0
37cos
120km
(making AC the subject of the formula)
= 150km (to 3 signicant gures)
Exercise 11j
1. A ship is on a bearing 060
0
from a lighthouse. What is the
bearing of the lighthouse from the ship?
2. A ship is on a bearing 200
0
from a lighthouse. What is the
bearing of the lighthouse from the ship?
3. P is the point due west of a harbour H and Q is a point
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which is 5km due south of H. If the distance PH is 7km, nd
the bearing of Q from P.
4. A boat leaves a harbour H on a bearing of 120
0
and it sails
100km on this bearing until it reaches a point B. How far is
B east of A? What distance south of A is B?
5. X is a port due west of a point P. Y is a point due south of P.
If the distances PX and PY are 10km and 15km respectively,
nd the bearing of X from Y.
6. A ship sails 35km on a bearing of 040
0
.
a) How far north has it travelled?
b) How far east has it travelled?
7. A ship sails 200km on a bearing of 240
0
.
a) How far south has it travelled?
b) How far west has it travelled?
8. An aircraft ies 400km from point O on a bearing of 025
0
and then 200Km on a bearing of 080
0
to arrive at B.
a) how far north of O is B?
b) how far east of O is B?
c) Find the distance and bearing of b from o.
9. An aircraft ies 500 km on a bearing of 100
0
and then
600km on a bearing of 160
0
. Find the distance and bearing
of the nishing point from the Starting point.
Unit summary
• In this unit you have learnt calculating angles and sides
of right angled triangles using trigonometric ratios. You
have also learnt how to derive fractional trigonometric
ratios of 30
0
, 45
0
, 60
0
and 90
0
. You also worked out problems
involving bearing and solved practical problems involving
trigonometry.
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Glossary
Tangent ratio: the ratio of the side opposite to a given angle in
a right angled triangle to the side adjacent that angle.
Sine ratio: the ratio of the side opposite to a given angle in a
right angled triangle to the hypotenuse.
Cosine ratio: the ratio of the side adjacent to a given angle in a
right angled triangle to the hypotenuse.
Special angles: angles whose ratios that can be expressed as
surds or simple fractions.
Unit review exercise
1. In each of the following triangles nd the sine of angle P :
Give your answers correct to 4 decimal places:
(a)
P
10.1 cm
20.5 cm
Q
R
(b)
Q
P
7.5m
1 0.9m
R
2.
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249
A
C 10 cm
B
35
0
In the above triangle, nd the length of AB.
3. A surveyor stands at the end of the bridge across a river.
The bridge is 20m long. The surveyor looks down at an
angle of 68
0
to see the bottom of the river on the opposite
side. Calculate the depth of the river to the nearest metre.
4. A 200m tower is to be built for relaying cellular phone
signals. The tower is to be anchored by cables from the top
of the tower that will each form a 65
0
angle with the ground.
Find how far from the base of the tower each cable will be
anchored.
5. The longest ladder that a City Council re department has
is 415cm mounted onto the roof a truck. For the safety of
those on the ladder, the re department does not want to
extend the ladder to an angle greater than 75
0
with the roof
of the truck. If the roof of the truck is 300cm off the ground,
nd the highest point the ladder can reach.
6. Calculate the lengths of the labeled sides in the triangles
below leaving your answers in surd form where possible:
(a)
50cm
x
30
0
(b)
x
25cm
60
0
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251
(c)
y
5
2
cm
45
0
7. A plane ying at an attitude of 10500m is to land at an
airport 25Km away. At what minimum angle must the pilot
lower his plane?
8. A ship left port X and travelled 70Km on a bearing S40
0
E to
port Y. It then travelled 100Km on a bearing S28
0
W to port
Z. Calculate
a. the shortest distance from X to Z correct to three signicant
gures.
b. the bearing of Z from Y to the nearest degee.
9. The bearing of ship H from town X is 145
0
36´ and from
town Y is 055
0
34´. Given that XY =10Km and that X is due
north of Y, nd the distance from H to Y correct to one
decimal place. = 5.7km
10. An aircraft ies 1300 km on a bearing of 210
0
and then
700km on a bearing of 060
0
. Find the distance and bearing
of the nishing point from the starting point.
References
R. Chikwakwa et al, Senior Secondary Mathematics (2002),
Macmillan, Malawi.
L. Bostock et al, National Curriculum Mathematics(1999),
Stanley Thornes, Cheltenham
Elain Ryder et al, CHANCO Teach yourself Mathematics(2013),
Chancellor College Publications, Malawi.
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Unit
12
SIMILARITY
In unit 9 of book 2, you studied
similar gures. Recall that similar
shapes are an enlargement of
each other. This means the length
of the larger shape are found by
multiplying the scale factor by the
lengths of the smaller shape.
In this unit, you will learn to apply
the ratio of areas and volumes of
similar gures to calculate areas,
sides of similar gures and volumes
of similar solids.
Similarity is used in many
situations. It is used in solving
real life problems such as nding
lengths, volumes, and areas of
similar objects such as triangles.
Ratio of areas of similar gures
Activity 1:
Identifying a scale factor of
similar gures
1. In pairs discuss the meaning of
“scale factor.”
2. Identify the scale factor in
the diagrams below ( A is the
object and B is the image)
9cm
3cm
4cm 12cm
3. Write the formula for nding
scale factor.
Scale factor shows how many times
an object has been enlarged. It is
found by the formula scale
Factor =
When the scale faclvtor is greater
than 1, the image is larger than the
object, when it is less than 1, the
image is smaller than the object and
when it is 1 the image and the object
are equal in size. A negative scale
factor upsides down the image.
Example 1:
Scale factor
The height of a triangle ABC is 10cm and
the height of a similar triangle A’B’C’ is
14cm. Calc ulate the scale factor.
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253
Solution
Scale factor =
14
10
= 1.4
Exercise 12a:
Calculate the scale factor in each of the following pairs of
gures. A is the object and B is the image:
1.
8cm
2cm A 4cm
1cm B
2.
7cm
A 14cm B
3. The actual length of a line segment is 2.5 m but it is
represented on a scale drawing by a line segment 2.5 cm.
Calculate the scale factor.
4. A distance of 1km is represented on a map by a line 2cm
long. nd the Scale factor used.
5. The radius of a circle is 14cm. If the circle is enlarged by the
scale factor of ½, nd the radius of the corresponding circle.
Activity 2:
Finding area factor of similar gures
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253
Working in pairs, discuss the meaning of area factor. Suggest
how area factor can be found.
The area factor of similar gures is the ratio of areas of similar
gures. It is found by the formula
Area factor =
Area of the image gure
Area of the object
Example 2
The area of a triangle is 24cm
2
. The area of another triangle an
enlargement of the rst triangle is 72 cm
2
. Find the area factor.
Solution
Area factor =
72
2
24
2
= 3
Exercise 12b:
In each of the following, A is the area of the object and B is the
area of the image . Find the area factor for each one of them:
a. A = 32 cm
2
and B = 48cm
2
b. A = 100 cm
2
and B = 50 cm
2
c. A = 54 cm
2
and B = 90 cm
2
d. A = 1.5m
2
and B = 1m
2
e. A = 100km
2
and B = 75km
2
f. A = 7.5 m
2
and B = 2.5m
2
1. Calculate the ratio of areas of the following pairs of solids.
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255
a.
2 cm
2 cm
8 cm
8 cm
2
c
m
8
c
m
b.
9cm
3cm
4cm 12cm
2. A rectangular garden 120m by 100m. Another rectangular
garden is150m by 120m. Find the ratio of the area of the
rst garden to that of the second garden.
3. The two parallel sides of a trapezium are 10cm and 20cm.
the perpendicular distance between them is 8cm. Another
trapezium has Parallel sides measuring 15cm and 30cm
with a perpendicular distance of 12cm between them. Find
the ratio of area of the second trapezium to the area of the
rst trapezium.
4. Calculate the ratio of area of two circles with radii 7cm and
21cm respectively
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255
Activity 3:
Calculating areas of similar shapes
In the previous three activities you have learnt how to identify
scale factor, area factor and how to calculate ratios of areas of
similar solids. In this section you will learn how to calculate
areas of similar shapes. But before you calculate areas of these
similar shapes, you need to know the relationship between scale
factor and area factor.
In groupsDraw any three pairs of similar gures of your choice.
Show the dimensions of he gures.
1. For each pair nd the scale factor and the area factor and
complete the table like the one below:
Scale factor Area factor
First pair
Second pair
Third pair
2. What is the relationship between the ratio of the lengths of
three pairs of similar shapes above and the ratio of the areas?
3. Report your ndings to class.
You must have found out that the area factors are the squares of
the scale factors or the scale factors are the square roots of the
area factors.
Example 3:
Areas of triangle
The two triangles, ABC and XYZ are similar. Given that area of
ABC is 5 cm
2
, nd the area of
∆
XYZ.
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257
A
C
B
4 cm
X
Y
Z
8 cm
Solution
Scale factor =
8
4
=
2
1
=
2
1
2
--------------------------
Relationship between scale factor and area
factor
=
Area of triangle XYZ
5cm
2
=
4
1
------------- Substitute area of triangle
ABC for 5cm
2
Area of triangle XYZ = 20cm
2
---- After cross multiplication
Exercise 12c
In this exercise, a number written inside a gure represents the
area of the shape in cm
2
. The number on the outside gives linear
dimensions in cm. In question 1 to 10, nd the unknown area A.
In each case, the shapes are similar.
1.
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257
6 cm
2
3 cm
9 cm
A
2.
3 cm
18 cm
5 cm
2
A
3.
10 cm
2
2 cm
6 cm
A
4.
7cm
21cm
A
1386cm
2
5
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259
20 cm
A
10 cm
50 cm
2
6
A
2 cm
8 cm
72 cm
7
10 cm
40 cm
2
A
15 cm
8
A
225 cm
2
8 cm
12 cm
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259
9
16 cm
2
L M
9 cm
18 cm
B
A
C
Find the area of LMCB.
Sometimes, you have the ratio of two similar shapes and you
are required to nd the ratio of the lengths. To do this, you must
take the square root.
Example 4
Two similar shapes have area 10cm
2
and 40cm
2
respectively. If
the length of the smaller shape is 6 cm, nd the corresponding
length in the larger shape.
Solution
Area factor =
40
2
10
2
=
4
1
Scale factor
Hence =
4
1
=
2
6
------- Area factor, scale factor relationship
=
4
1
=
2
36
--------- Remove the brackets
x
2 =
144 --------- Cross multiply
x = 12cm ------- Take square root of both sides
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261
Exercise 12d
In questions 1– 6 nd the lengths marked for each pair of
similar shapes.
1.
cm
2 cm
x
3 cm
2
12 cm
2
2.
cm
6 cm
2
54 cm
2
18 cm
x
3.
cm
400 cm
2
100 cm
2
20 cm
x
4.
cm
20 cm
2
45 cm
2
x
5. P and Q are regular pentagons. Q is an enlargement of P
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261
by a scale factor 3. If the area of pentagon Q is 180 cm
2
,
calculate the area of P.
6. The rectangular oor plan of a house measures 8 cm by 6
cm. If the scale of the plan is 1:50,
Calculate:
a) the dimension of the actual oor
b) the area of the actual oor in m
2
.
7. A garden has an area of 3025 m
2
, and is represented on a
plan by an area of 144 m
2
. Find the actual length of a wall,
which is represented on the plan by a line 8.4 m long.
2. Ratio of volumes of similar solids
Having studied area factor and scale factor in the previous
sections, you will now study the volume scale factor of
similar gures and its application.
Activity 4:
Finding the volume scale factor of similar solids
1. Working in pairs, discuss the meaning of “ volume scale
factor”.
2. Suggest how volume scale factor can be found.
3. Find the volume scale factor of the models and the drawn
gures.
The volume scale factor of similar gures is the ratio of volumes
of similar gures. It is found by the formula
Volume factor =
Volume of the image gure
Volume of the object
You will later on learn in the coming sections the relationship
between scale factor and volume scale factor.
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263
Example 5
The volume of a container A is 300cm
3
and the volume of
container B is 150cm
3
. Given that container B is an enlargement
of container A, nd the volume scale factor of the two containers.
Solution
Volume scale factor =
150
3
300
3
=
1
2
Exercise 12e
1. Calculate the volume scale factor for each of the following
pairs of Volumes. A is the volume of the object and B is the
volume of the.
a. A = 250cm
3
; B = 100cm
3
b. A = 240cm
3
; B = 320cm
3
c. A = 64m
3
; B = 8m
3
d. A = 576m
3
; B = 1728m
3
e. A = 244cm
3
; B = 61cm
3
f. A = 17.4m
3
; B = 11.6m
3
g. A = 350m
3
; B = 1050m
3
h. A = 112cm
3
; B = 28cm
3
2. Calculate the ratio of volumes of the following pairs
of similar solids.
The gure to the right is the image
a.
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263
1.4cm 2.8cm
b.
6cm
3cm
15cm 7.5cm
12cm
6cm
Activity 5:
Calculating of volumes of similar gures
In the previous section you have learnt how to nd and calculate
the ratios of volumes of similar solids. You will now learn to
calculate volumes of similar solids and to nd lengths of sides
of similar solids using the relationship between scale factor
and volume factor. You will rst nd the relationship between
volume scale factor and scale factor by doing the following
activity:
Working in groups, perform the following activity
1. Find the volume of the cube below:
2cm
2cm
2cm
2. Using other similar cubes, build the solid below:
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265
3. What is the length, width and height of the above solid?
4. Deduce the volume of the solid you have built above.
5. Build two more solids by adding one cube along the length,
the width and the height each time to the preceding solid.
Find the volumes of the resulting solids.
6. Now complete the table below:
Length of a side Volume of the solid
Solid 1 2 cm 8 cm
3
Solid 2 4 cm 6 4cm
3
Solid 3
Solid 4
Table 1
7. Now try dividing the lengths and volumes of any two cubes
as follows:
Solid 1 and solid 2 : Ratio of sides: 2cm = 1
4cm 2
Ratio of volumes: = (8 cm
3
) = 1
(64 cm
3
)
8
Do the same with Solids 2 and 3 and Solids 3 and 4.
8. Present your work by lling the table below:
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265
Scale factor Volumes factor
Solids 1 and 2
Solids 2 and 3
Solids 3 and 4
Table 2
9. What is the relationship between the ratio of volume of
similar gures and their sides?
You might have discovered that the volume factor of similar
solids is the cube of the scale factor or you may also say
that the scale factor of similar gures is the cube root of the
volumes factor s.
Example 6
Calculate the ratio of volumes of similar cylinders of diameter 6
cm and 9cm.
Solution
Let the volume of the smaller cylinder = x cm
3
and that of a
larger
Cylinder = ycm
3
So x cm
3
: ycm
3
= 6
3
: 9
3
i.e x : y = 216 : 729
= 8 : 27
Example 7
Two spheres have volumes in the ratio 64:125. What is the ratio
of their surface areas?
Solution:
The ratio of corresponding sides of the spheres =
64
125
3
=
3
√64
3
√125
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267
The ratio of their surface areas =
4
2
5
2
= 16:25
Example 8
Two similar cylinders have their radii in the ratio 1 : 2. If the
smaller cylinder has a volume of 21.56m
3
, calculate the volume
of the larger cylinder.
Solution
Since the volume factor of similar solids is the cube of scale
factor of the similar solids, or since the scale factor of similar
solids is the cube root of the ratio of the volume factor,
1
2
=
21.56
3
1
3
2
3
=
21.56
-------Cube both sides of the equation.
1
8
=
21.56
x = 172.8m
3
(After cross multiplication)
Exercise 12f
1. Calculate the ratio of volumes of similar rectangular tanks
with lengths 12cm and 16cm.
2. Two similar solids have their surface areas as 160cm
2
and
360cm
2
. Calculate the ratio of their volumes.
3. The volume of a tank 5m high is 343m
3
. Calculate the
volume of a similar tank 10m high.
4. Two similar solids have surface areas in the ratio 9 : 25. If
the volume of the smaller solid is 81cm
3
, what is the volume
of the larger solid?
5. A cylindrical tin 1.5m high is lled with 50m
3
of liquid and
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267
the liquid rises to a level of 1m. Calculate how much more
liquid the tank can hold so that it is completely full.
Unit summary
• In this unit you have learnt to calculate the ratios of areas
and volumes of similar gures. You have also learnt to apply
these ratios to calculate areas and sides of similar gures.
Unit review exercise
1. Two heaps of sand are both in a conical shape. The height
of a smaller heap is 70cm and the height of a bigger heap
is 56cm. If the smaller heap contain 128g of sand, calculate
the mass of the bigger heap.
2. The ratio of the radii of two similar containers is 3:2. The
larger container has a capacity of 4.05litres. Calculate the
capacity of the smaller container.
3. The height of a triangle is 6cm and its area is 12cm
2
.
Calculate base of a similar triangle whose area is 48cm
2
.
4. The areas of two similar parallelograms are in the ratio of
4:3. The height of the larger parallelogram is 8cm. Find the
height of the smaller parallelogram.
5. In a scale drawing of a building, a wall 1.5m high is shown
to be 10cm high. Calculate the actual width of the building
which is shown by a width of 5cm on the drawing.
6. The volume of a balloon is Vcm
3
and its radius is 6cm. When
the balloon is inated further its volume increases by 40%.
Find the new radius of he balloon.
7. A container is partially lled with 200cm
3
of water and the
water rises to a height of 50mm. calculate the volume of
water that must be added to increase the depth by 20mm.
8. The fuel tank of a truck 1m long is in a rectangular form.
When completely full the tank can hold 405litres of fuel.
Calculate the length of similar tank for a small lorry that
has a capacity of of the truck.
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269
Glossary
Scale factor: The ratio of the length of a side of an image to the
corresponding side of a similar object.
Area factor: The ratio of the area of the image to the area of
the object.
Volume factor: Ratio of volumes of similar gures.
References
S. Hau and F. Saiti (2002), Strides in Mathematics Book 3,
Longman , Malawi
Larson etal (1998), Heath Algebra an Integrated Approach,
Heath and company, Canada.
G. D. Buckwell and B.N Githua, Gold Medal Mathematics,
Macmillan, London.
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269
Unit
13
COORDINATE GEOMETRY
You are familiar with locating
a point in a plane and using
coordinates to describe the
position of a point in a plane.
These coordinates are called the
Cartesian coordinates of the point.
(The name comes from the French
mathematician Rene Descartes
(1596 −1650). The system of using
a pair of coordinates to describe
the position of a point in a plane
is called Coordinate Geometry or
Cartesian Geometry.
The coordinates measure the
displacement (+ or −) of the point
from two perpendicular axes, the
y-axis (Oy) and the x − axis (Ox)
where O is the origin.
In this unit, you are going to learn
how to solve problems involving
straight lines in an x − y plane.
You will learn how to calculate the
distance between two points on a
straight line, how to nd the equation
of a straight line, and nally you will
learn to describe the condition for two
lines to be parallel and how to nd
the midpoint of a line segment.
Coordinate Geometry is applied in
many situations. For example by
recognising the relationship between
two variables, you can learn more
about real life situations and make
reasonable predictions about future
trends such as those concerning
populations and business.
Calculating the distance
between two points on a
straight line
We can represent the position of
a point with respect to the y axis
and x axis by two numbers called
coordinates. It is for this reason
that the xy-plane is also called the
coordinate plane. These numbers
are enclosed in brackets and the rst
number in the brackets represents the
position of a point with respect to the
x-axis and the second number in the
brackets represents the position of a
point with respect to the y- axis.
If we want to nd the distance
between two points on the coordinate
plane, then a simple right angled
triangle can be constructed with sides
parallel to the axes.
270
271
Activity 1:
Calculating the distance between two points by using
Pythagoras theorem
Work in groups:
1. Using a scale of 2cm to represent 1 unit on both axes draw
the straight line whose equation is y = x + 2 for the values
of x =2 , 4 , 6
2. Using a ruler, measure the length of the line segment from
x=2 to x = 6.
3. Record your ndings.
Now on the same graph paper draw straight lines y = 4 and
x = 6. Let the two straight lines intersect at P.
4. What type of triangle is formed by the line y = x + 2 and the
two lines you have just drawn?
5. Measure the lengths of the two legs of the triangle. Can you
nd a way of nding these lengths by using the coordinates
on the two vertices of the triangle which are not right
angles?
6. Using Pythagoras Theorem, nd the length of the
hypotenuse of the triangle and compare your result with the
result you found in step 3.
7. Now generalize your results to nding the length of a
straight line from A(x
1
, y
1
) to B(x
2
,y
2
) where (x
1
, y
1
) and
(x
2
, y
2
) are two points through which the straight line
passes.
Example 1:
Distance between two points
Find the length of the line joining A (1, 2) and B (3, 4)
AB =
( ) ( )
22
2413 −+−
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271
=
22
22 +
= 8
=
22
= 2.83 (to 3 signicant gures)
Exercise 13a
Find, correcting your answers to 2 decimal places where
necessary, the length of the line joining
1. A (1, 2) and B (4, 6)
2. A (4, 2) and B (2, 5)
3. A (3, 4) and B (0, 0)
4. A (−1, −3) and B (2, 1)
5. A (−4, −5) and B (1, 7)
6. A (0, −3) and B (4, 0)
7. A (−1, −3) and B (−2, −5)
8. A (−2, 1) and B (4, 2)
9. A (−5, −2) and B (0, −3)
10. A (−5, 0) and B (−7, −4)
The equation of a straight line
The equation of a straight line is a relationship between two
variables x and y. You drew equations of straight lines in Form
2 when you were solving simultaneous equations graphically. In
this section you shall continue to look at these lines in the xy-
plane.
272
273
Activity 2:
Writing the equation of a straight line in the form
y = mx + c
Generally, there are two forms of the equation of a straight line.
These are ax + bx = c and y = mx + c. In the rst form, a, b
and c are numbers. In the second form m is the gradient of the
straight line and c is the y intercept.
In groups,
1. Write your own examples of equations on each of the above
forms.
2. Present your answers on the chalkboard.
3. Together with your teacher, group the answers you have
presented into the two equation forms.
4. Group the following equations into the two forms in 1
above:
a. 3y = 2x +1
b. 2y – 3x + 8 = 0
c. y = 2x
d. 4x + 3y = 6
e. 2x –1 = y
All the equations that have y on one side of the equation is
said to be in slope intercept form. Note that in this form the
coefcient of y must be 1.You should also have seen that the
basic idea in writing the equation of a straight line in slope
intercept form is to make y the subject of the equation.
Example 2
Write the equation y + 2x = 3 in the form y = mx + c
Solution
y = -2x + 3 ------------- moving 2x from the left side to the right
side of the equation.
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273
Example 3:
Equation of a line
Write the equation 3y – 5x + 9 = 0 in slope intercept form.
Solution:
Move -5x and +9 to the right side of the equation:
3y = 5x – 9
Divide by 3 throughout:
y = x – 3
Exercise 13b
Write the following equations of straight lines in slope intercept
from:
(1) 2y + 2x = 5
(2) 3y = 6x – 13
(3) 5y + 3x – 22 = 0
(4) 2x – y = 9
(5) x + 2y + 6 = 0
(6) x + y = 6
(7) 8x + 3y = 48
(8) 2y – x +1 = 0
Activity 3:
Relating gradient to the tangent of an angle
Consider the triangle below:
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275
P
20cm
40°
Q
R
1. Express tan 40
0
as a ratio of the two legs of triangle PQR.
2. Find the length of QR; give the answer correct to one decimal
place.
3. Find the ratio
PQ
QR
to 3 decimal places.
4. Now nd tan 40
0
using a calculator.
5. Compare the results in 3 and 4. What conclusion do you draw
from the results?
You might have noted that the gradient of a straight line PR is
equal to the tangent of 40
0
or the tangent of its complement i.e.
50
0
.
Example 4
Equation of a straight line
A straight line makes an angle θ with the x - axis. The line cuts
the x axis at x = 8. If tan θ = ¼, nd the equation of the straight
line in slope intercept form.
Solution
Substitute m for ¼ into y = mx + c
y = ¼ x + c
When y = 0, x = 8 so 0 = ¼ (8) + c
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275
0 = 2 + c
c = -2
The equation is y = ¼ x -2
Activity 4:
Using the relationship between gradient of a line and
tangent of an angle to nd gradient of a straight line
In Activity 3, we have seen that the gradient of a line and the
tangent of an angle (other than 90
0
) in a right angled triangle
are numerically equal. As long as the lengths of the legs of the
triangle are known, we can nd the gradient of the straight line.
Consider the triangle below:
6 P
1 Q R
0 2 4
1. Discuss how you can nd the lengths of PQ and QR without
having to use distance formula.
2. Use your ndings to nd the lengths of PQ and QR.
3. Hence nd the gradient of line PR by dividing PQ by QR.
leave the gradient as a fraction with a numerator and
denominator only.
4. Write the coordinates of P and R and discuss how you can
use the coordinates to nd the result in 3 above.
5. Generalise the result in 4 above with P(x
1
,y
1
) and R(x
2
,y
2
).
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277
The gradient of a straight line is found by dividing the difference
between the two y-values by the difference between the two x-
values. If the gradient is 0 the line is parallel to the x axis. The
straight line parallel to the y-axis has an undened gradient.
Example 5:
Gradient of a straight line
Find the gradient of a straight line passing through (3,6) and
(8,3).
Solution
Gradient =
3 6
8
3
=
3
5
Exercise 13c
Find the gradient of a straight line passing through each of the
following pairs of points:
1. (2,3) and ( 6,9)
2. (4, 11) and ( 2, 5)
3. ( 2.5,3.5) and ( 8.5 , 9.5)
4. ( 1, 2) and ( 6,5)
5. ( 4, 3) and ( 2, 8)
6. ( 4.5 , 5 ) and ( 2 , 10)
7. ( p , 2p) and ( 3p, 5p)
8. (2x,3y) and ( x , -2y)
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277
Activity 5:
Formulating the equation of a straight line with a given
gradient and through a given point
We looked at this method when we solved example 3 on page
272. Carefully study this example again and rephrase the
question to contain the word “tangent” and the phrase “passing
through”.
Compare your rephrased question with those from other groups.
You can use the slope intercept form or the point intercept
form of a line to formulate the equation of a straight line with a
given gradient and through a given point. This is illustrated in
example 6 below:
Example 6:
Equation of a straight line
Find the equation of a straight line passing through (3, 5) and
with gradient .
Solution
Using slope intercept form,
Substitute y for 5, x for 3 and m for
2
/
3
into y = mx + c:
5 = (3) + c
∴
5
= 2 + c
∴
5
– 2 = c
∴
c = 3
∴
The equation is y =
2
/
3
x + 3
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279
OR
Using point intercept form,
y – y
1
= m(x-x
1
) where (x
1,
y
1
) is the given point and m is the
given gradient
x
1
=3 and y
1=
5
y – 5 =
2
/
3
(x – 3)
∴
y – 5 =
2
/
3
x – 2
∴
y =
2
/
3
x – 2 + 5
∴
y =
2
/
3
x + 3
Exercise 13d
Find the equation, in slope intercept form, of a straight line
passing through
1. (1,3) with gradient -3.
2. ( -2,-5) with gradient
3. (-3, 1) with gradient ½.
4. (7, 5) with gradient -2.
5. ( 1,-2) with gradient
6. (0,0) with gradient 1
7. ( 2,0) with gradient 3
8. (a, 2a) with gradient 2
Activity 6:
Finding the equation of a straight line passing through
two given points
1. On a graph paper, draw a straight line through A(-2 , 3) and B ( 4, -1)
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279
2. Using your knowledge for nding gradients of straight lines,
nd the gradient of line segment AB.
3. Now choose any point P on the straight line .Call this point
(x, y). Using this point and the coordinates of point A, nd
the gradient of line segment AP.
4. Do you think the gradients in steps 2 and 3 are different?
Come up with a relationship for x and y using these results.
5. Try to use the point P(x, y) and point B (4, -1) to nd the
relationship between x and y. Compare your result to the
result in step4 and comment on the result.
The relationship you found above is called the equation of the
line through A and B. As you have seen, to nd the equation
you only need to nd the gradient of the line (see Activity 4) and
any one of the two points through which the line passes. You can
then use either the slope intercept form or the point intercept
form to nd the equation of the line.
Example 7:
Equation of a straight line
Find the equation of a straight line passing through (1, 3) and
(4, 9)
Solution
First nd the gradient of the line m =
2
3
6
14
39
==
−
−
∴
The equation is of the form y = 2x + c.
To nd the value of c substitute, the values of x and y from
any one of the two points.
Using the rst point we have 3 = 2
×
1 + c
∴
c = 1
∴
The equation is y = 2x + 1
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281
Exercise 13e
Find the equation of the line passing through the points A and B
in the form y = mx + c when
a. A = (2, 4) and B = (3, 8)
b. A = (0, 2) and B = (3, 5)
c. A = (−2, 0) and B = (2, 8)
d. A = (3, −1) and B = (7, 3)
e. A = (−4, −1) and B = (−3, −9)
f. A = (0,0) and B = ( 2,3)
g. A = ( 3,5) and B = (1,1)
Activity 7:
Finding the equation of a straight line from the graph
Consider the graph below:
−1
1
1
0
2
2
3
3
4
4
Q
x
y
P
−1
−2
−2
Discuss how you can nd the equation of line PQ in the graph
above.
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281
To nd the equation of a straight line from the graph, you need
to obtain any two “smart points” on the line and use them to
nd the gradient of the line. Then the slope intercept or the
point intercept form can be used to nd the equation of the
straight line.
Example 8:
Finding equation of a line from a graph
=
=================
o======
ó
=
== =
P
=
=
== =
O
=
=
== =
N
=
=
J
O
=
N
=
M
=
N
=
O
=
P== = ====Q
=
ñ
=
=
= =
JN
=
=
= =
JO======== ======= ======= ======= ====
=
================ ======= ======= ======= ==========================
p
=
=
Find the equation of the straight line RS above.
Solution
Two smart points are (0, 2) and ( 3,-2)
Gradient =
2 2
0
3
= -
4
3
The equation is y - 2 = -
4
3
(x – 0)
y = -
4
3
x + 2
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283
-6 -5 -4 -3 -2 -1 0 1
8
7
6
5
4
3
2
1
→x
→
y
→
→
-6 -5 -4 -3 -2 -1 0 1
5
4
3
2
1
-1
-2
→x
→
y
→
→
-1 1 2
7
6
5
4
3
2
1
-1
-2
→x
→
y
→
→
-2 -1 0 1
3
2
1
-
-1
-2
-3
-4
-5
-6
→x
→
y
→
→
1.
2.
3.
5.
-1 1 2
7
6
5
4
3
2
1
-1
-2
→x
→
y
→
→
4.
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283
Exercise 13f
Find the equations of each of the following straight lines:
Parallel lines
In this section we shall look at parallel lines. You have used
the idea of parallel lines in many situations e.g. when you
were learning about properties of parallelograms in your
JCE Mathematics or when you are using alternate angle or
corresponding angle properties. But what makes lines parallel?
Let us investigate this question by looking at Activity 8.
Activity 8:
Gradients of parallel lines
In your JCE you learnt about drawing parallel lines.
1. On a graph paper, draw two straight parallel lines.
2. Choose two smart points on each line and nd the gradient of
each line.
3. Compare the gradients of the two lines. What do you notice?
In this activity you must have seen that parallel lines have the
same gradient.
Example 9:
Showing that lines are parallel
Show that the line passing through ( 1,3) and ( 2, 7) is parallel to
the line passing through (4,3) and (3,-1).
Solution
Gradient of the rst line =
7 3
2
1
= 4
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285
Gradient of the second line =
3 (-1)
3 4
=
4
1
= 4
As the gradients of the two lines are the same, the two lines are
parallel.
Example 10:
Equation of line parallel to another line
Find the equation of a line through the point (1, 2) which is
parallel to the line 2x – 3y = 4.
Solution
First, put the given equation of the line into standard form i.e.
y =
3
4
3
2
+x
∴
The line has gradient
3
4
This is also the gradient of the line whose equation we must nd
because the lines are parallel.
∴
Using slope intercept form,
y =
3
4
x + C
Substitute x = 1, y = 2 into this form,
∴
2 =
3
4
×
1 + C
∴
C =
3
4
∴
T
he equation is y =
3
4
3
2
+x
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285
Exercise 13g
1. Find the equation of the line which is
a) parallel to x – y = 1 and passes through (2, 3)
b) parallel to 2x + y = 3 and passes through (3, 0)
c) parallel to y + 5x = 2 and passes through (1, 3)
d) parallel to 2x – y = 4 and passes through (0, 3)
e) parallel to x – 3y = 1 and passes through (−2, −1)
2 A straight line passes through A(5,7) and B( 0,-1). Another
straight line passes through Q(-3,-5) and R(2,3). Show that
AB is parallel to QR.
3 Show that lines with equations 4x – 3y = 7 and
9y -12x - 1 = 0 are parallel.
4 Which of the following lines are parallel?
a. From ( -1,3) to (4,5)
b. From ( 2, -1) to ( 7,1)
c. From ( 0,-1) to ( 2,2)
Midpoint of a line segment
The midpoint of a line segment is a point which lies halfway
between the end points of a line segment. It is possible to nd
the midpoint of a line if we know the coordinates of the end
points of the line. We rst derive the formula for nding the
midpoint of a line segment.
Activity 9:
Deriving the formula for the mid-point of a line segment
1. On a graph paper and using 1 cm to represent 1 unit on the
horizontal axis, plot the points A (2, 6) and B (6, 3).
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287
2. Use the distance formula to nd the length of line AB and
then get half of the length. Note that your answer is in cm.
3. From point A and along line AB measure the number of cm
you found in step 2 above and write down the coordinates of
the point you nd.
4. What is the connection between the coordinates in 3 and the
coordinates in 1?
5. Come up with a generalization for nding the midpoint of a
line segment using (x
1
, y
1
) and ( x
2
,y
2
) as two points on the
straight line.
You have seen from this activity that the midpoint of a line
segment is found by nding half of each of the sums of the x
coordinates and the y coordinates i.e.
Midpoint =
2
1
(x
1
+ x
2,
y
1
+ y
2
)
Example 11:
Mid-point of a line
Find the midpoint of a line from (2, 6) to (4, 8)
Solution
The midpoint is
(
2+8
2
,
6+8
2
)
= (5, 7)
Example 12
The midpoint of line PQ is (4, 9). The coordinates of P are (a, 5)
and of Q are (1, 7). Find the value of a.
Solution
(
+
,
+
)
= (1, 7)
So
+
= 1
4 + a = 2 ------------- multiplying both sides by 2
= -2
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287
Exercise 13h
1. Find the midpoints of the lines from:
a. (3, 4) to (1, -5)
b. (-1, -4) to (-8, -10)
c. (0, 3) to (2, -5)
d. (p, 2p) to (3p, -4p)
2. The midpoint of BC is (2, 4). If the coordinates of B are ( 1,8)
nd the Coordinates of C.
3. The coordinates of the point of intersection of the diagonals
of a parallelogram ABCD are (3.5, 2.5). A is the point (2, 5)
and B is a point (8, 8). Find the coordinates of C and D.
4. Find the distance between the midpoint of a line from (2, 2)
to (4, 5) and a point (7, 9)
5. U(2, 4) , V(5, 4) , W (1, 4) and X(0, 6) are four points on a
Cartesian plane. Find the gradient of the line joining the
mid points of lines UV and WX.
6. The equation of a straight line is 2x + 3y = 7. Find the
equation of a line passing through the midpoint of A(8, 6)
and B(2, 4) parallel to the line 2x + 3y = 7.
Solving real life problems using coordinate geometry
The knowledge of coordinate geometry is useful many areas. The
following few examples show how useful coordinate geometry
can be in solving real life problems:
Example 13:
Real life problem using coordinate geometry
A contractor is to build an ofce for a company. He however
has to build the ofce in such a way as to suit the physically
challenged. Given that the recommended maximum slope for a
wheelchair ramp is and the contractor is to build a ramp that
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289
rose to a height of 2m, calculate the minimum horizontal length
of the ramp.
Solution
The fraction
1
/
10
means y increases by 1m for every 10m increase
in x. You can model an equation as follows:
y
( 0,1)
x=
(10,0)
The equation of the straight line is y - 1= (x – 0)
y - 1 = x
y = x + 1
To nd the minimum length of the ramp, put y = 2
2 = x +1
2 – 1 = x
1 = x
x = 10
This means the minimum length of the ramp should be 10m.
Example 14:
Real life problems using coordinate geometry
During the 2009 general elections in Malawi, 43 women were
elected into parliament. In 2014, the number of elected women
dropped to 30.Write a linear model for the number y, of elected
women who were parliamentarians between 2009 and 2014.
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289
Solution
Let x = 0 represent 2009. You can draw the following xy-plane
y
43
30
x
2009 2014
Gradient of the straight line =
43 30
2009
2014
=
13
5
Y intercept is 43,so the equation is y =
-
13
5
x + 43
Unit summary
• In this chapter you have learnt to nd the distance between
two points on a straight line, how to nd the equation and
gradient of a straight line and how to nd the midpoint of a
line segment. You have also learnt to write the equation of a
straight line in slope intercept form and to nd the gradients
of parallel lines.
Glossary
Smart point: A point on the xy-plane where the straight line
graph passes through a vertex of a grid box.
-
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291
Unit review exercise
1. Find the length of a line joining the following points:
(a) A( 3,8) and ( 3,3)
(b) P( -3,4) and Q ( 0, 6)
2. Write the following equations of straight lines in slope
intercept form:
(a) 3x – 2y = 13
(b) 4x +2y – 7 = 0
3. Find the equation of a straight line
(a) passing through (4, 5) and ( 1, 2).
(b) passing through ( -2,4) with gradient
4. The vertices of a triangle are A(-1, 8) , B(-2 , 0) and C(0, 0).
Show that triangle ABC is isosceles.
6. The line y = mx + 4 passes through the point (1 , 2).
Calculate
a. the value of m
b. the angle that this line makes with the y – axis.
c. the coordinates of the point where this line cuts the
x – axis.
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291
Unit
14
VARIATIONS
In your JCE Mathematics you learned about
direct and inverse proportion. You learnt
that direct proportion is the relationship
between related quantities such that they
both increase or decrease in the same ratio.
You also learnt that inverse proportion is
the relationship between quantities such
that as one quantity increases (decreases)
the other decreases (increases) in the same
but opposite ratio. In this book, these
proportions will be referred to as direct
variation and inverse variations. You will
learn to solve problems involving direct
and inverse variations. You will also
solve problems involving joint and partial
variations.
The knowledge of variation is used in many
real life situations. For example, among
others, it can be used to nd the number of
people required to do certain work, it can be
used to nd the speed at which a car must
travel so as to cover a certain distance in a
given time and it can be used in physics to
nd the force required to lift an object and
in many other situations.
Direct variation
Two related quantities are said to
be in direct variation if they have
the same rate regardless of the
variables. For example if the ratio
of y: x = k then you can write y = kx
or if you multiply both sides of this
equation by x then you have y = kx.
The number k is called the constant
of variation.
Activity 1:
Modeling direct variation
In pairs, discuss how you can model
direct variation from the following
statement:
The speed, s (in km per hour), of a car
varies directly with time, t (in hours).
After the car has travelled for 2 hours
its speed is 40km/h. write a model
that gives speed of the car in terms
of t.
Compare your work with the other
pairs.
Let your teacher check your work.
Example 1:
Modeling direct variation
The exchange rate of Malawian
Kwacha (K), to the United States
Dollar ($) is 1: 475.
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293
If the exchange rate is in direct variation, write a model that
gives K in terms of ($). Solution
K = c ($), c is constant of variation
1 = 475c ............. Substitute K for 1 and ($) for 475
c = 1 .............. Divide both sides by 475
475
K =
1 .
475
($)
Exercise 14a
1. If y varies directly as x, write a model that expresses y in
terms of x when y = 2 and x = 4.
2. The pressure (p) of an object submerged under water is
directly proportional to the depth (h). When p = 460, h = 5.
Model an expression that gives p in terms of h.
3. The number of Calories, c, a person can burn and the time,
t(in minutes), the person spends on doing activity vary
directly. 150kg people can burn 75 Calories by sitting in a
class for 50 minutes. Write a linear model that relates c and
t.
4. The perimeter, p, of a square is proportional to the length,
l. When l = 4, p =16. Create a linear model that give p in
terms of l.
5. The number of bags (n) is directly proportional to the cost
(C in Kwacha) per bag. If 200 bags cost K40000, model a
linear expression which expresses n in terms of C.
Activity 2:
Deriving the general equation involving direct variation
In section 1.1 you modeled expressions representing direct
variation. You used the general form y = kx. In this section you
shall learn how this form can be derived. In groups go through
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293
the following activity:
1. Suppose a 50kg bag of maize costs K5000. Construct a table
of number of bags (n) against total cost(c) for up to 5bags.
2. What happens to the total cost as the number of bags
increases?
3. Pick any two numbers of bags and nd their ratio in
simplest form and then pick their corresponding costs and
nd their ratio in simplest form. How do the two pairs of
ratios compare?
4. Try other pairs in your table in a similar manner and
comment on the results.
5. Now try dividing corresponding number of Kwachas by
the number of bags for all the entries giving the ratios in
simplest form. Comment on your ndings.
You have seen that as the number of bags increases, the total
cost increases in the same ratio. You can then say that the
number of bags (n) is directly proportional to the total cost(c).
The symbol for variation is
∝
. So you can write
c
∝
n to mean “c
is directly proportional to n” or “c varies directly as n” or
“c varies as n.”
Additionally, you have seen that the ratio
c
/
n
is the same for all
the corresponding entries or
c
/
n
= constant. Hence if n
∝
c
then
c
/
n
= k where k is a Constant called the constant of variation.
Challenge:
Find your own example of a direct variation where the
quantities decrease in the same ratio.
Example 2:
Direct variation
Given that y varies directly as x and that y =3 when x = 6 nd
the value of y when x = 10.
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295
Solution
Since y varies directly as x then = k or y = xk where k is a
constant
∴
3 = 6k------------------substitute y for 3 and x for 6
∴
k = ½ -------------------divide by 6 both sides.
∴
the law of variation is y = ½ x
∴
When x = 10, y = ½ (10) i.e. y = 5
Example 3:
Derect variation
The mass (m) of each piece of log of wood that can be cut from
the same log of wood is directly proportional to the length (l) of
the piece. A piece 30cm long has a mass of 2kg. What will be the
mass of a 45cm long log of wood?
Solution
If m
∝
l then m = kl
∴
2 = 30l
∴
l =
1
45
The law of variation is m =
1
15
l
when l = 45, m =
1
15
(45)
m = 3kg
Exercise 14b
1. If d
∝
t and d = 80 when t = 5, nd d when t = 3.
2. p
∝
q and p = 4.5 when q = 12. Find p when q = 16.
3. The time swing, t seconds, of a pendulum clock varies as the
square root of its length, l, cm. If t = 1 when l = 25 calculate
l when t = 1.5 seconds.
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295
4. The cost (c) per metre of a cloth is proportional to the
number of metres (n) bought. If 2metres cost K750.00,
(a) Find the relationship between c and n
(b) Hence nd n when c = K3750.00
5. The cost, c, of painting a wall is directly proportional to the
area, A, to Be painted. An area of 300m
2
costs K3500. Find
the cost of painting 210m
2
6. In a given period, the cost, (C), of paying for accommodation
at a hotel varies as the number (n) of participants present.
The cost of paying for 60 participants is K18000. Find the
cost for paying for the accommodation of 100 participants.
Activity 3:
Presenting direct variation graphically
The models you modeled in section 1.1 can be presented
graphically. You rst develop a table of values, plot them on a
graph paper and then draw the graph of the model. In groups,
perform the following activity:
1. Model a linear expression for a quantity y which varies
directly as x given that x = 20 when y = 10.
2. Using the model, nd the corresponding values of y when
x = 0, 2, 4, 6, 8, 10.
3. Present your values in a table form.
4. Using the values in the table, draw the graph for the
variation.
5. What type of graph is produced?
Graphs of direct variations are straight line graphs. The slope is
represented by the constant of variation.
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297
Example 4:
Direct variation
P  Q and P = 20 when Q = 100, nd the relationship between P
and Q and sketch the graph of the relation P
∝
Q.
Solution
Since PQ, P = kQ, where k is a constant.
So, 20 = 100k ---- Substitute P for 20 and Q for 100
k =
20
100
k =
1
5
The relationship is P =
1
5
Q
To develop the table for the relation use multiples of 20
( P-value in the given question) for P. One of the ways is shown
below:
P 0 20 40 60 80 100
Q 0 4 8 12 16 20
Then draw the graph as follows:
100
80
P
60
40
20
0
4
8
12
16
20
Q
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297
Exercise 14c
For each of the following, nd the relationship between the
variables and draw the graphs for the relationships:
1. y ∝ x and y = 40 when x = 20
2. y ∝ x and y = 8 when x = 32
3. P ∝ Q and P = 20 when Q= −10
4. R ∝ T and R = 10 when T = −10
5. L ∝ T and L = 15 when T = 5
6. M ∝ N and M = −30 when N = 10
7. V ∝ W and V = 40 when W = 200
Inverse variation
So far you have studied direct variation. You will now study
one more type of variation called inverse variation. Variables
x and y are said to be in inverse variation if the product of the
two variables is constant i.e. xy = k, where k is the constant of
variation.
Activity 4:
Modelling inverse variation
In pairs, discuss how you can model inverse variation from the
statement below:
The number of days, (d), it takes to build a wall varies inversely
as the number, (n), of boys available. If there are 20 boys the
wall takes 10 days to build. Model an expression that gives d in
terms of n.
Compare your work with the other pairs.
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299
Example 5:
Inverse variation
The time (t), taken by a car to cover a given distance varies
inversely as speed, (S in km/h). At 60km/h, the car takes 3hours.
Model an expression that expresses t in terms of S.
Solution
t S = k, where k is a constant --------- Model for inverse variation
3 (60) = k ----- Substitute t for 3 and S for 60
180 = k ------- Simplify
So tS =180 ------- Substitute k for 180.
Exercise 14d
1. Model an inverse variation equation for each of the
following for the given values of the variables:
(a) P varies inversely as Q. (P = 0.025, Q = 0.04).
(b) I varies inversely as R
2
. (I = 10, R = 1).
(c) T varies inversely as the square root of d. (T = 3, d = 64).
(d) P varies inversely as q (s = 0.5, q = 8).
2. The time (t in seconds) taken to type a 2000 word document
varies with the rate (r) at which you can type. When the
rate is 40 words per minute, you can nish typing the
document in 50 minutes. Write a model that relates t and r.
3. The distance (d metres) required to balance a seesaw is
inversely proportional to ones weight (w in kg). A 120kg
can balance the seesaw at 60cm away from the centre of the
seesaw. Model an inverse variation expression that relates d
and w.
4. Prepare your own questions on modeling inverse variation.
Let your teacher check them.
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299
Activity 5:
Deriving the general equation involving inverse
variation
Having looked at modelling inverse variation, you will now
learn how the general equation for this type of variation can be
derived. In pairs, go through the following activity:
1. Suppose a motorist is to cover 20km. Construct a table of
speed (s) against time (t) showing the time the motorist will
take to cover the distance if she travels at 5km/h, 10km/h
and 20km/h respectively.
2. What happens to the time taken as the speed increases?
3. Pick any two numbers of kilometres and nd their ratio in
simplest form and then pick their corresponding numbers of
speeds and nd their ratio in simplest form. How do the two
pairs of ratios compare?
4. Try other pairs of numbers in your table in a similar
manner and comment on the results.
5. Now try multiplying the number of kilometres by the
corresponding number of hours for all the entries. Comment
on your ndings.
You have seen that as speed increases, the time it takes to cover
the distance of 20km decreases in the same but opposite ratio.
You can then say that the “time (t) is inversely proportional to
speed( s)” or “time (t) varies inversely as speed (s)”. You
write t
∝
1
/
s
to mean “time (t) is inversely proportional to speed (s)” or
“time (t) varies inversely as speed (s)”.
Additionally, you have seen that the product ts is the same for
all the corresponding entries or ts = k where k is a constant.
Hence if
t
∝
1
/
s
then ts
= k where k is a constant called the
constant of variation.
Challenge
Can you nd your own examples of inverse variation where
one quantity decreases and the other increases in the same but
opposite ratio.
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301
Example 6:
Inverse variation
Given that y
∝
1
/
x
and that when y = 60, x = 12. Find the value
of x when y = 25.
Solution
If y
∝
1
/
x
then xy = k, k constant
∴
12 × 60 = k
∴
k = 720
∴
The law of variation is xy = 720
∴
When y = 25, then 25x = 720
∴
x = 28.8
Example 7:
Inverse variation
The number of days (d) it takes to complete to paint a wall is
inversely proportional to the number (n) of men available to do
the work. 12men take 15 days to paint the wall. How long will
18men take to paint the wall?
Solution
If d
∝
then dn = k, k constant
∴
12x15 = k
∴
k = 180
∴
The law of variation is dn = 180
∴
When n = 18, then 18d = 180
∴
d = 10
∴
18 men will take 10 days to paint the wall.
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301
Exercise 14e
1. If y
∝
1
/
x
, and that when y = 100, x = 25. Find the
relationship between x and y hence nd x when y = 20.
2. Given that f is inversely proportional to d and that when f =
35, d = 7. Find d when f = 25.
3. q varies inversely as t. When q = 2, t = 8. Find the value of t
when q is increased by 80%.
4. The time (t in minutes) it takes a typist to nish typing a
document is inversely proportional to the number of words
(w) she types per minute. if she types 250 words per minute
she takes 30minutes. How long will it take her to nish
typing the document if she types 100words per minute?
5. The effort (e in Newtons) applied to lift a load placed 100cm
from the fulcrum is inversely proportional to the distance
(d in cm) away from the fulcrum on the other side. An
effort of 200N is required to lift the load when the distance
is 60cm. What effort will be needed to lift the load if the
distance is 40cm?
Activity 6:
Presenting inverse variation graphically
In this section you will learn how you can present the models of
inverse variation graphically. In your groups
1. Draw the table of values for the model yx = 100 using the
values x = 0, 5, 10, 15, 20, 25.
2. Using the values and a scale of 1cm to represent 5 units on
the x – axis draw the graph of the equation xy = 100.
3. Present your work to the other groups. let also your teacher
check your work.
You have seen that the graph of inverse variation generally
looks like this:
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303
y
x
Exercise 14f
Rewrite xy = k starting with y and draw the same graph with
negative values of x.
3. Joint variation
Joint variation is a variation whereby one variable varies with
the product of two or more other variables. For example, if
z = kxy, where k is a constant, then z varies jointly with the
product of x and y. You will now learn how to derive the general
equation involving joint variation.
Activity 7:
Deriving the general equation involving joint variation.
In form 2 you learnt the formula for volume of a cylinder. In
pairs,
1. Write down the formula.
2. Does the formula depend on the size of a cylinder?
3. Write the formula as a direct variation.
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303
4. What is the value of the constant of variation?
You have seen that the formula for the volume of the cylinder
applies to all the cylinders. The volume depends on the square of
the radius and the height i.e. V r
2
h i.e V = kr
2
h where k =
22
7
.
Hence the general equation of a joint variation can be
derived from a direct variation.
Example 8:
Joint variation
p
∝
qr. When q = 2 and r = 5, p = 50
(a) Find the equation connecting p,q and r
(b) Find p when q = 5 and r = 4.
Solution (a)
=9
qr
∴
kqrp =
where k is a constant
Substitute
2,50 == qp
and
9
729d
in the equation above
∴
5250 ××= k
∴
k1050 =
∴
9
729d
∴
qrp 5=
This is the equation connecting
qp,
and
d
(b)
qrP 5=
Substitute
9
729d
and
9
729d
∴
455 ××=p
=162
∴
729=k
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305
Example 9:
Joint variation
m varies directly as d and inversely as the square of t. if
729=k
when
9
729d
and
9
729d
, nd:
(a) m when
9
729d
and
9
729d
(b) d when
72=m
and
9
729d
Solution
(a) m
∝
d and m
∝
2
1
t
(b)
=M
2
729
t
d
∴
m
∝
2
t
d
∴
=72
2
3
729 d×
∴
m =
2
t
kd
where k is a constant.
∴
=72
9
729d
∴
2
3
2
162
×
=
k
∴
d729972 =×
∴
=162
6
4k
∴
d
=
729
972×
∴
k21629 =×
=
81
72
∴
=9
4
3729 ×
=
9
8
∴
729=k
= 0.9 (to 2
decimal
places)
∴
2
729
t
d
m =
=
4
3729 ×
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305
=
=9
75.546=
Example 10:
Joint variation
m varies directly with the square of n and inversely with p.
when n=2 and p=6, m=9. Find m when n=4 and p=8.
Solution
m
∝
6
4k
∴
=m
p
kn
2
∴
=9
6
2
2
×k
∴
=9
6
4k
∴
k454 =
∴
=k
4
54
2
27
=
or 13.5
∴
The law of variation is
m =
27n
2
2p
∴
When n = 4 and p = 8, m=
27(4
2
)
2(8)
∴
m= 27
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307
Exercise 14g
1. y varies directly with x and z. when x = 3 and z =10, y =15
Find the relationship between x, z and y.
2. a
∝
b
2
c. when a = 100, b = 2 and c = 5
(a) nd a when b = 6 and c = 5
(b) nd b when a=72 and c = 6.4
3. p
∝
R
Q
when Q = 75, R = 25 and p = 125
(a) nd the equation connecting p, Q and R
(b) Find p when Q = 3 and R =100.
4. x, y and z are related quantities such that x varies directly
as y and inversely as the square of z. When x =24 and y =
3, z = 4
(a) Find the value of x when y = 4 and z = 8
(b) Find the value of z when x =128 and y =
4
1
.
5. r varies directly with t and inversely with s. r =27 when t =
9 and s = 2. Find r when t = 7 and s = 6.
6. If x
∝
y and y
∝
z
. How does x vary with z?
7. P
∝
qr and q varies inversely with the square of r. How does
p vary with r?
8. x varies directly with y and z. When y = 6 and z = 3, x = 7
2
1
.
Find (a) x when y =12 and z = 5
(b) Z when x =2.5 and y = 4
9. The mass m of a cylindrical tin varies jointly with its height
(h) and the square of its radius (r). If the tin 50 cm high
and of radius 2cm has a mass of 540g; Find the mass of a
cylindrical tin 10cm high and of radius 4cm.
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307
4. Partial variation
This is the type of variation which consists of two or more
parts added together. The relationship in these parts varies
from question to question. Since there are two or more
parts added together, there are at least two constants which
are used in solving these questions. Hence the knowledge
of simultaneous linear equations is essential at this stage
since there is need to solve for the unknowns which are used
for constants in the equations.
Partial variation usually applies in situations where the
value of one major quantity is the sum of two or more
quantities varying in different ways. For example the
cost of running a boarding school depends on two separate
factors and these are overheads, such as electricity, water,
telephone bills, and wages of employees and on the other
hand the cost of food used in the feeding of students in the
school. In most cases, overheads are constants i.e. the bills
and wages are still paid without considering the fact that
there are students in the school or not. The cost of food is
the only quantity, which is proportional to the number of
students being fed at the school. Therefore, we can say that
the total cost is partly constant and partly varies as the
number of students.
Activity 8:
Formulating the general equation of a partial variation
You have so far learnt to formulate the general equation of
direct, inverse and joint variations. You will now learn to
formulate the general equation of a partial variation.
Look at the equation y = 3 + 2x. In pairs, come up with your own
examples of such equations. Discuss how similar the equations
you have given are.
You might have noted that the right hands of all the equations
are made up of two components: one is a direct variation and
the other is a constant. So if y = 3 + 2x, you can say y is partly
constant and partly varies as x. You should also note that in this
equation and in the equations you gave in activity 8, there are
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309
two constants. You can therefore write y = a + kx as a general
equation of partial variation where a and k are constants of
variation.
Note: The knowledge of simultaneous linear equations is
essential at this stage since there is need to solve for the
unknowns which are used for constants in the equations.
Example 11:
Partial variations
A is partly constant and partly varies with B. When B = 2, A =
10 and when B = 4, A = 16
(a) Find the relationship between A and B
(b) Find A when B = 5.
Solution:
(a) From the rst sentence A = a + b B where a and b are
constants.
From the second sentence 10 = a + 2b (i) and 16 = a + 4b (ii)
Subtract (i) from (ii)
16 = a + 4b
− 10 = a + 2b
6 = 0 + 2b
6 = 2b
∴
b = 3
Substitute b = 2 in equation (i) to nd b
10 = a + (2 x 3)
10 = a + 6
∴
a = 10 – 6
= 4
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309
Hence A = 4 + 3B is the required formula. (from A = a + bB)
(b) When B = 5
∴
A = 4 + 3 × 5
= 4 + 15
= 19
Example 12:
Partial variation
A varies partly as C and partly as the square root of C. When C
= 4, A = 22 and when C = 9, A = 42
(a) 4
Solution
(a) From the rst sentence A = aC + b
C
where a and b are
constant
From the second sentence
22 = 4a + b
4
(i)
And 42 = 9a + b
9
(ii)
i.e. 22 = 4a + 2b (i)
And 42 = 9a + 3b (ii)
Equation (i) x3 66 = 12a + 6b (iii)
Equation (ii) x2 84 = 18a + 6b (iv)
Subtract (iii) from (iv)
84 = 18a + 6b
− 66 = 12a + 6b
18 = 6a + 0
18 = 6a
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311
∴
a = 3
Substitute a = 3 in equation (i) to nd b
22 = 4 x 3 + 2b
22 =12 + 2b
10 = 2b
∴
b =5
Hence A = 3C + 5
c is the required equation.
(b)
when C = 25
A = 3 × 25 + 5
25
= 75 + 5 × 5
= 75 + 25
= 100
Exercise 14h
1. P is partly constant and partly varies as Q. When Q = 3,
P = 22 and when Q =2, P = 18.
(a) Find the equation connecting P and Q.
(b) Find P when Q =10.
2. x is partly constant and partly varies as the square of y.
When y = 2, x =20 and when y = 3, x = 25; nd the value of x
when y = 7.
3. The cost of making a sofa set is partly constant and partly
varies with the time it takes to make. If it takes 3 days to
make, it costs K27, 000. If it takes 5 days to make, it costs
K31, 000; nd the cost if it takes 2 days to make?
4. The cost of running a boarding school is partly constant and
partly varies as the number of students in the school. If
there are 100 students, the total cost is K45,000 per month
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311
and if there are 240 students the total cost comes to K87,
000 per month. Find the total cost if there are 360 students
in the school.
5. M varies partly as N and partly as the square root of N.
When N = 4, M = 6 and when N = 16, M = 16. Find M when
N = 25.
6. E is partly constant and partly varies with S. When S = 40,
E = 50 and when S = 54, E = 92. Find S when E = 32.
7. The cost of producing mathematics books is partly constant
and partly varies as the number of books produced.
The total cost of producing 4 books is K2500 and that
of producing 10 books is K5800. Find the total cost of
producing 1000 books.
8. The cost of transporting people in a minibus from Blantyre
to Lilongwe is partly constant and partly varies as the
number of people transported. Transporting 15 people costs
K8750 and transporting 20 people costs K10, 000. Find the
cost of transporting 10 people within the same distance.
9. P is a quantity, which varies partly as V and partly as V
2
. When
V = 4, p =52.8 and when v =5, p = 81. Find p when v = 3.
10. The resistance to motion of a bicycle is partly constant and
partly varies as the square of the speed. At 8 km/hr, the
resistance is 80N and at 12km/hr, the resistance is 100N.
Find the resistance if the speed of the bicycle is 20km/hr.
Activity 9:
Presenting partial variation graphically
You are familiar with equations of the form y = mx + c.
In pairs,
1. Compare this form with the general equation involving
partial variation.
2. From your comparison, what type of graph is a partial
variation?
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313
You must have seen that partial variation graphs are straight
lines in slope intercept form.
Example 13:
Partial variations
y is partly constant and partly varies as x. When x = 10, y = 2
and when x = 4, y = 14. Express y in terms of x and draw the
graph of the variation.
Solution
y = a + kx ---------- General partial variation equation.
2 = a +10k (i) ---------- Substitute y for 2 and x for 10
14= a + 4k (ii) ---------- Substitute y for 16 and x for 4
−12 = 6k ----------------------- Subtract (ii) from (i)
k = −2 ------------------------- Divide by 6 both sides
2 = a − 2 (10) ---------------- Substitute k for -2 into equation(i)
2 = a – 20
a = 22 ------------- Rearrange
The expression is y = 22 – 2x
The graph is a straight line whose y – intercept is 22 . The graph
also cuts the x axis at −11 (i.e when y = 0)
The graph is drawn below:
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313
y
22
y = 22 -2x
x
011
Exercise 14i
Draw partial variation graphs for the following questions:
1. A quantity y is partly constant and partly varies as x. When
y = 5, x = 2.5 and when y = 3 , x = 1.5.
2. p is partly constant and partly varies as q. When p =3, q = 1
and when p = 6 , q = 2.5
3. A is partly constant and partly varies as B. When A = 11, B
= 7 and when A = 15 , B = 9.
4. The cost (C in Malawi Kwacha) of making an item is partly
constant and partly varies with time(t in minutes) it takes
to make the item. When t = 100, C = K498 and when t = 30,
C = K148.
5. T is partly constant and partly varies as C. When C =5,
T = 37 and when C = 15, T = 57.
Unit Summary
• In this unit you have learnt direct, inverse, joint and partial
variations. You have learnt to model direct and inverse
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315
variations and to derive general equations involving the four
variations. You also learnt how you can present variations
graphically and solve variation problems.
Unit review exercise
1. If y varies directly as x, write a model that expresses y in
terms of x when y = 9 and x = 3.
2. Q is directly proportional R . When Q = 100 , R = 5. Model
an expression that gives Q in terms of R.
3. If y
∝
x and y = 80 when x = 5, nd y when x = 3.
4. V
∝
W and V = 4.5 when W = 12. Find V when W = 16.
5. Given that y
∝
and that when y = 20, x = 10. Find the value
of x when y = 25.
6.
Given that y
∝
and that when y = 2, x = 3. Find the value
of x when y = 1.
7
. If T varies directly as the cube of S and that T = 6 when S =
2, nd T when S = 8.
8. The mass (m in grams) of a solid varies directly as the
volume(V in cm
3
) of the solid. The mass is 12g when the
volume is 8 cm
3
. Find the volume for which the mass is 30g.
9. The resistance , R , of a wire varies inversely with the cross
sectional area of the wire, A. A 500cm wire length copper
wire with a radius of 0.1cm has a resistance of 0.1ohms.
nd the resistance of the same length of copper wire with a
radius of 0.2cm.
10. The intensity, I in watts per m
2
, of jet engine noise at
Kamuzu International Airport varies inversely as the
square of the distance, r in m
2
. At a distance of 1 metre the
intensity is 10 watts per square metre. An airport cargo
worker is 15metres from the jet engine. Calculate the
intensity of the jet noise at that distance.
11. The royalties, in Kwacha, that an author receive on the
sale of a book are partly constant and partly varies as
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315
the number of books sold. On sales of 10000 she receives
K80000 and on the sales of 6000 she receives 50000. Find
the amount she will receive on the sales of 15000copies.
12. The cost of producing a component is partly proportional
to the area of the component and inversely proportional to
the number of components produced each day. On a day
120 components of area 10cm
2
were produced and the cost
was K100 per component. On the other day 180 components
of area 20cm
2
were produced and the cost per component
was K80. Find the cost per component of producing 50
components per day of area 50cm
2
.
Glossary: None
References
S. Hau and F. Saiti (2002), Strides in Mathematics Book 3,
Longman , Blantyre
Larson etal ( 1998),Heath Algebra An Integrated Approach,
Heath and company, Ottawa.
G. D. Buckwell and B.N Githua, Gold Medal Mathematics,
Macmillan, London.
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317
Unit
15
GRAPHS OF QUADRATIC
FUNCTIONS
Recall from unit 1 that a quadratic
function is a function of the form
ax
2
+ bx + c where a, b and c are
constants and a ≠ o.
Youlearnt how to factorize quadratic
expressions and how to solve
quadratic equations by factorisation,
completing the square and by the
quadratic formula.
In this unit, you are going to learn
about the graphs of quadratic
functions and the properties of such
graphs. You shall learn how to draw
and interpret graphs of quadratic
functions. You shall also learn how to
solve quadratic equations and linear
and quadratic equations graphically
and how to formulate quadratic
equations given quadratic graphs
which cut the x – axis.
The knowledge of quadratic graphs
is used in many situations such as
in the study of falling objects and in
studying quantities related to time
among others.
Drawing graphs of quadratic
functions
Activity 1:
You are familiar with completing or
drawing a table of values of a given
equation from your JCE Course.
In groups,
1. Discuss how you construct a
table of values for any given
equation.
2. Now construct a table of values
for the equation y = x
2
for −4 ≤
x ≤ 4.
3. Using a scale of 2cm to
represent 1 unit on both axes,
plot the points on the graph.
What shape have the points
formed?
4. Using a free hand draw the
graph of y = x
2
through all the
points.
5. Repeat steps 1, 2 and 3 but
now use the equation y = −x
2
.
6. Comment on the similarities
and differences between the
two graphs.
You should have noted that if the
graph of a quadratic function is
plotted and drawn, a smooth curve
is produced. The curve is called a
parabola. The parabola is either cup
shaped when the coefcient of x
2
is positive or inverted(cap shaped)
when the coefcient of x
2
is negative.
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317
Exercise 15 a
For each of the following quadratic functions, construct a table
of values and then draw the graph.
1. y =
2
1
x
2
for −4 ≤ x ≤ 4
2. y = −
2
1
x
2
for
−4 ≤ x ≤ 4
3. y = 2x
2
for
−4 ≤ x ≤ 4
4. y = −2x
2
for
−4 ≤ x ≤ 4
5. y = x
2
+ 1
for
−4 ≤ x ≤ 4
6. y = x
2
− 1
for
−4 ≤ x ≤ 4
7. y = −x
2
− 1
for
−4 ≤ x ≤ 4
8. y = x
2
− x − 2
for
−4 ≤ x ≤ 4
9. y = −x
2
+ 2x + 3
for
−3 ≤ x ≤ 5
10. y = 3x
2
− 3x − 6
for
−2 ≤ x ≤ 3
Interpreting graphs of quadratic functions
In this section, you shall learn to interpret graphs of quadratic
functions. You shall learn to describe the effect of a, b and c
on the nature of graph of y = ax
2
+ bx + c, nd maximum and
minimum values of a quadratic function and to nd the equation
of the line of symmetry.
Activity 2:
The effect of changing the value of a in y = ax
2
In Activity 1 you drew the graph of y = x
2
. You will now see more
closely what happens to this graph as the value of a change.
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319
In groups,
1. Draw the graphs of y = x
2
, y = 2x
2
, and y = 3x
2
on the same
axes and using the same scale.
2. How do the three graphs compare to the graph of y = x
2
?
3. What is the major difference amongst the graphs?
4. Write down the coefcients of the three graphs in order of
their steepness starting with the less steep.
5. What can you say about the effect of the change in the value
of a on the graph of y = x
2
?
You have seen that as the value of a decrease, the graphs
become more and more open or they become less and less
steep. In all the three graphs, the line x = 0 divides each graph
into equal halves. In other words, the line x = 0 is the line of
symmetry.
Activity 3:
The effect of adding or subtracting a constant from the
quadratic squaring function i.e. the equation y = x
2
In your groups,
1. Draw on the same axes and the same scale, the graphs of
y = x
2
, y = x
2
− 1and y = x
2
+ 1 for values of x from −4 to +4.
2. What are the similarities and the differences between the
graphs?
3. How do the shapes of the two graphs compare to that of
y = x
2
?
4. What can you say is the effect of adding or subtracting a
constant to x
2
?
You have seen from this activity that all the three graphs have
the same shape and that in all the three graphs, the line x = 0 is
the line of symmetry. The three graphs only differ in the points
where they cut the y axis. This enables you to come up with the
following interpretation about the graphs:
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319
(a) Adding a constant to the quadratic squaring function i.e
to x
2
shifts the graph vertically upwards along the line of
symmetry. The graph shifts by the value of the constant.
(b) Subtracting a constant from the quadratic squaring function
i.e. to x
2
shifts the graph vertically downwards along the
line of symmetry. Again, the graph shifts by the value of the
constant.
(c) The constant also gives the minimum value of the quadratic
function for cup shaped parabola or maximum value for cap
shaped parabola.
Example 1:
Sketching graphs
Draw sketches of the following graphs on the same system of
axes using the same scale on both axes:
a. y = x
2
+ 4 b. y = x
2
- 3
Solution
From activity 3,
• The graph of y = x
2
+ 4 has the shape of y = x
2
but shifted
vertically upwards by 4 units and that y = 4 is the minimum
value of the function.
• The graph of y = x
2
- 3 has the shape of y = x
2
but shifted
vertically downwards by 3 units and that y =-3 is the
minimum value of the function.
• x = 0 is the line of symmetry
The graphs are drawn below:
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321
y = x
2
- 3
y
x
-3
4
0
y = x
2
+ 4
Activity 4:
The effect of adding or subtracting a constant from x and
then squaring the result
Again in your groups,
1. Draw, on the same axes and scale, the following graphs:
y = x
2
, y = (x − 1)
2
and y = (x + 2)
2
for values of x from −4 to
+4.
2. What are the similarities and the differences between the
graphs?
3. What is the equation of the line of symmetry for the graphs?
4. What can you say is the effect of adding or subtracting a
constant from x and then squaring the result?
You have seen that the graph of y = (x − 1)
2
has the same shape
as the graph of y = x
2
but has been shifted to the right by 1
unit. The graph of y = (x + 2)
2
also has the same shape as the
graph of y = x
2
but has been shifted to the left by 2 units. The
lines of symmetry are the lines x = 1 and x = 2 respectively. The
minimum value of each function is 0. The square of the constant
gives the y-intercept.
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321
You can therefore make the following interpretations about
adding or subtracting constants to x before squaring:
(a) Adding a constant to x before squaring shifts the graph
of y = x
2
horizontally to the left by the value – constant
(minus the given constant).
(b) Subtracting a constant from x before squaring shifts the
graph of y = x
2
horizontally to the right by the value positive
constant (plus the given constant).
(c) The minus constant and the plus constant are also the new
lines of symmetry.
(d) Y-intercept is the square of the constant.
Example 2:
Sketching graphs
On the same axes draw sketches of the graphs of y = (x − 3)
2
and
y = (x + 4)
2
Solution
From activity 4,
• The graphs of y = (x − 3)
2
and y = (x + 4)
2
are the same as
the graph of y = x
2
but shifted horizontally to the right by
3 units and to the left by 4 units respectively.
• 0 is the minimum value of each function.
• x = 4 and x = 3 are the lines of symmetry respectively.
The graphs are drawn below:
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323
y
y = ( x -3)
2
y = ( x + 4)
2
9
x
-
4 0 3
=
16
Example 3:
Drawing graphs
Draw a sketch of the graph of y = x
2
– 2x + 3
Solution
You need to relate this function to the forms y = ax
2
or y = (x − a)
2
rst. These are the functions that you have looked at in activity
3 and 4. You do this by writing the given function into a perfect
square plus a number by adding and subtracting (coefcient of
x)
2
as follows:
y = x
2
– 2x + {½(–2)}
2
– {½(–2)}
2
+ 3
y = x
2
– 2x + 1 – 1 +3
y = x
2
– 2x + 1 + 2
i.e. y = (x – 1)
2
+ 2
You can see that expressed in this form, y = x
2
– 2x + 3 is the
graph of y = x
2
shifted in two directions: 1 unit to the right
and then 2 units upwards or in short, it is the graph of y =
(x – 1)
2
shifted upwards by 2 units. The equation of the axis of
symmetry (which is also the x –value at the turning point of the
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323
graph) is x = 1 and the minimum value of the function is
y = 2.the graph is drawn below together with the graphs y = x
2
and y = (x – 1)
2
for you to see their relationship:
y= x
2
– 2x + 3
y
y= x
2
y = (x – 1)
2
3
2
1
0 1
x
You will now investigate the effect of increasing or decreasing
the values of a , b and c in the equation y = ax
2
+ bx + c .
Activity 5 :
Finding the effect of increasing or decreasing the values
of a , b and c in the equation y = ax
2
+ bx + c .
Copy the following table in your exercise books before you go
into your groups. Fill in this table as you go through the activity:
Increasing
the value
a
Decreasing
the value of a
Increasing
the value
of b
Decreasing
the value
of b
Increasing
the value
of c
Decreasing
the value
of c
Effect on
turning
point
Effect on
line of
symmetry
1. In your groups, draw the graph of y = x
2
– 2x – 8, clearly
showing the line of symmetry and the turning point of the
graph. Note that in this equation a = 1, b = − 2 and c = − 8.
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2. Now, keeping the values of b and c constant, increase the
value of a and draw the graph of the resulting equation on
the same axes.
3. Show the equation of the line of symmetry and the turning
point of this graph.
4. How do the two compare with their positions in the rst
graph?
5. Increase the value of a even further and repeat the process.
6. Fill the rst column of your table.
7. Repeat the above steps and ll column 2 by decreasing the
value of a.
8. Investigate the effects of b and c in a similar manner and
complete the table. Note that as you investigate the effect of
b, a and c must be kept constant and when you investigate
the effect of c, a and b must be kept constant.
Investigate the effect of a on one graph the effect of b on another
graph and similarly the effect of c.
Challenge
Use the equation of a cap shaped parabola to investigate the
effects of a , b, c on the nature of the parabola. Come up with a
table similar to the one you used on the cup shaped parabola in
activity 5.
Exercise 15b
Draw sketches of the following graphs. In each case, state the
equation of the axis of symmetry and the coordinates of the
turning point of the function:
1. y = x
2
− 2
2. y = −x
2
+ 1
3. y = −(x − 1)
2
4. y = 3x
2
+ 6x
5. y = 2x
2
+ 5x + 2
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Activity 6:
Finding the maximum and minimum values of a
quadratic function
A minimum or maximum value of a quadratic function is
the value of y at the turning point of the graph. Earlier on
in activity 1 of this unit, you saw that a quadratic graph is
either cup shaped or cap shaped. A cup shaped parabola has a
minimum point i.e. lowest point while the cap shaped parabola
has a maximum point i.e. highest point. In short, a cup shaped
parabola has a minimum value while cap shaped parabola has
a maximum point. Each of the two points lies on the line of
symmetry. You will now learn how to nd the maximum and
minimum values of a quadratic function by doing two group
activities.
In groups,
1. Draw up a table of values for the quadratic functions y = x
2
+ 2x − 3 and 4 +3x – x
2
for −4 x 4.
2. Using the tables, draw the graph of the quadratic function y
= x
2
+ 2x − 3 and 4 + 3x – x
2
3. What is the value of x at the minimum or maximum point
from the graphs?
4. Now, nd the roots of the two quadratic functions and
complete the table below:
Equation of parabola Roots of the
parabola
x value at the turning
point of the parabola
x
2
+ 2x – 3
4 + 3x – x
2
5. Now nd the relationship between the roots and the value of
x at the turning Points of each graph.
You must have seen that the value of x at the turning point of
the graph is half the sum of the two roots.
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Example 4:
Turning points
Find the x value at the turning point of the graph
y = 2x
2
– 7x + 3
Solution
First nd roots of the equation 2x
2
– 7x + 3 = 0.
(x – 3) (2x – 1) = 0 ------------ By factor method
x = 3 or x = ½
Hence the x vale at the turning point is ½ (3 + ½) = 1.75
Exercise 15c
Find the value of x at the turning points of the following graphs:
1. y = x
2
− 5x + 6
2. y = x
2
− 4x – 5
3. y = x
2
+ 8x + 15
4. y = 4x
2
− 5x – 6
5. y = 4x
2
− 3x −10
Having seen the relationship between the roots and the value
of x at the turning point of a graph, you will now see the
relationship between the x value and the constants a and b of a
parabola ax
2
+ bx + c.
Activity 7
In groups, study the table below which shows the roots of
parabolas, the turning points, the x value at the turning points
and the values of a and b in the parabola. (One fraction in
column 3 has deliberately not been simplied)
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Equation of parabola Roots of the
parabola
x value at the turning
point of the parabola
Coefcients of x
2
and x
y = x
2
– 8x + 15 3 , 5
8
/
2
a = 1 and b = –8
y = 6x
2
– 7x + 2
1
/
2
,
2
/
3
7
/
12
a = 6 and b = –7
y = 3x
2
– 13x + 12
4
/
3
, 3
13
/
6
a =3 and b = -13
y = 6x
2
+ 7x – 5
1
/
2
,
–5
/
2,
–7
/
12
a = 6 and b = 7
Discuss the relationship that you see between the x value at the
turning point and the constants a and b of a parabola in each
row. Write down this relationship. Use your own equations and
verify that the relationship is true.
You must have seen that in column 3 the numerator is the
value of b with opposite sign and that the denominator is twice
the value of a. The results can be generalised as follows: the
x – value at the turning point ( maximum or minimum) of the
parabola y = ax
2
+ bx + c, where a, b and c are constants is
x =
a
b
2
−
and this also gives the equation of the line of symmetry.
To nd the corresponding value of y, substitute x =
a
b
2
−
into the
quadratic equation. This gives the maximum or minimum value
of the quadratic function.
Example 5:
Maximum and minimum points
For the function y = x
2
+ x − 6, what is
(i) The value of x at the minimum point?
(ii) The equation of the line of symmetry?
(iii) The minimum value of y.
(i) at the minimum x =
a
b
2
−
, and for this quadratic function
a = 1, b = 1, c = −6
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∴
x = –
2
1
12
1
−=
×
.
(ii) The equation of the line of symmetry is x =
1
2
(iii) y =
4
1
66
2
1
2
1
2
−=−
−+
−
.
Exercise 15d
Find the maximum or minimum values of the following
functions and the equation of the line of symmetry.
1. y = x
2
+ 2x – 6 2. y = 2x
2
+ 3x – 5
3. y = x
2
– 4x + 1 4. y = x
2
+ 2x – 3
5. y = 1 − 2x – x
2
6. y =4 – 8x – 2x
2
7. y = −3x
2
− 12x + 1 8. y = 5 − 2x – 4x
2
9. y = (x – 1)
2
10. y = (2x + 1)
2
4. Solving quadratic graphs graphically
You learnt to solve quadratic equations in unit 1 of this
book. In that unit you used factorisation and quadratic
formula. You will now learn to solve quadratic equations
graphically.
Activity 8:
Finding roots of a quadratic equation ax
2
+ bx + c = 0
graphically
In your groups,
1. Draw the table of values for the function y = x
2
+ x – 6 for
the values -4 x 4.
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2. Using a scale of 2cm to represent 1 unit on both axes draw
the graph of y = x
2
+ x – 6 on the graph paper.
3. Read off the values of x at the two points where the graph
cuts the x axis i.e where the graph cuts the line y = 0.( note
that since y = 0, you have actually drawn the graph for the
equation 0 = x
2
+ x – 6 or x
2
+ x – 6=0)
4. Now solve the equation x
2
+ x – 6 = 0 by factors and compare
your result with the values you found in 3 above. What do
you nd?
It can be seen from the results that to solve the equation ax
2
+ bx +c = 0, the quadratic graph is drawn and the roots of the
equation are the x values at which the graph cuts the x-axis.
Exercise 15e
Solve each of the quadratic equations below by plotting a graph
for −4 ≤ x ≤ + 4. On the horizontal axis, use a scale of 1cm to
represent 1 unit. On the vertical axis, use the scales indicated in
the brackets against each question:
1. x
2
– x – 6 = 0 ( 2cm to represent 5units)
2. –x
2
+ 1 = 0 ( 2cm to represent 5units)
3. x
2
– 6x + 9 = 0 ( 2cm to represent 10units)
4. – x
2
– x + 12 = 0 ( 2cm to represent 5units)
5. x
2
– 4x + 4 = 0 (2cm to represent 10units)
6. 2x
2
– 7x + 3 = 0 ( 2cm to represent 20units)
7. x
2
+ 3x – 10 =0 ( 2cm to represent 10units)
8. 3x
2
– 5x – 2 = 0 ( 2cm to represent 20units)
9. 2x
2
– 7x = 0 ( 2cm to represent 20units)
10. x
2
– 4x + 3 = 0 ( 2cm to represent 10units)
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Activity 8:
Solutions to simultaneous linear and quadratic equations
In activity 7, you solved quadratic graphs by nding the points
of intersection of a quadratic graph and any horizontal line. You
will now learn how to solve graphically equations of the form
ax
2
+ bx + c = d, where d is less than 0 or is greater than 0. You
will also learn to solve quadratic equations by nding the points
of intersection of the quadratic graph and other linear equations
of the form y = mx + c.
Suppose you want to solve graphically the equation x
2
+ 3x − 6 =
4 for −6 x 3
1. Draw up the table of values for the equation y = x
2
+ 3x – 6
for −6 x 2.
2. Using a scale of 2 cm to represent 1 unit on both axes draw
the graph of y = x
2
+ 3x − 6
3. Now draw the line y = 4 on the same axes as y = x
2
+ 3x – 6.
4. Read off the values of x at the points where the two graphs
intersect.
5. Now solve the quadratic equation x
2
+ 3x − 6 = 4 by factors
and compare the results with the results they found in 3.
Comment on your ndings.
You have learnt from the above activity that provided that
solutions exist, all quadratic expressions equated to a constant
can be solved by drawing the graphs of y = the quadratic
expression and y = the constant and their point(s) of intersection
give the solution to the quadratic equation.
Some equations that can be solved in a similar way are those
of the form ax
2
+ bx + c = ax + c where a, b and c. Go through
activity 9 below:
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Activity 9:
Graphical solutions by points of intersection of the
quadratic graph and a linear expression of the form
y = mx = c
Suppose you want to solve graphically the equation
x
2
+ 2x – 1 = x + 1 for −6 x 3.
In groups,
1. Using the values of x in the given range, draw up the tables
of values for the two equations.
2. Using a scale of 2cm to represent 1 unit on both axes draw
the graph of y = x
2
+ 2x − 1 and y = x + 1
3. Read off the values of x at the points where the two graphs
intersect.
4. Now solve the quadratic equation x
2
+ 2x − 1 = x + 1 by
factors by grouping and compare the results with the results
you found in 3. What do you nd?
You have seen that the results in 3 and 4 are the same.
This means that provided that solutions exist, all quadratic
expressions equated to a linear expression can be solved by
drawing the graphs of y = the quadratic expression and y =
the linear expression and their point(s) of intersection give the
solution to the quadratic equation.
Example 6:
Graphical solutions
Solve graphically the equation x
2
+ x – 6 = 2x + 3
Solution
In this case you draw the graphs of y = x
2
+ x – 6 and y = 2x + 3
on the same pair of axes.
y = x
2
+ x – 6
x −4 −3 −2 −1 0 1 2 3 4
y 6 0 −4 −6 −6 −4 0 6 14
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333
For graphs of straight lines, you only need three pairs of points.
y = 2x + 3
x −3 0 3
y −3 3 9
Now you can draw the two graphs
The two graphs intersect when x = −2.5 and when x = 3.5. Hence
the roots of the equation x2 + x – 6 = 2x + 3 are x = −2.5 and x =
3.5.
Sometimes you are asked to draw a graph and use it to solve
other quadratic equations.
Example 7:
Grapical solutions
(a) Draw the graph of y = 3x
2
– x – 2 for −3 ≤ x ≤ 3.
(b) Use the graph to solve the following equations:
( i) 3x
2
– x = 12
(ii) 3x
2
– 7 = 0
(iii) 3x
2
= 2x + 5
y=x
2
+x-6
y=2x–6
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333
(iv) 2 + x – 3x
2
= 0
Solution notes:
Draw the graph of y = 3x
2
– x – 2 as in the previous activities.
Then proceed as follows to answer part (b) of the question:
i. Rearrange 3x
2
– x = 12 to get 3x
2
– x − 12 = 0. Then draw
the graph of y = 3x
2
– x – 2 – (3x
2
– x –12) i.e. y = 10. The
roots of the equation 3x
2
– x = 12 are the x values at the
points of intersection of the two graphs.
ii. As before, draw the graph of y = 3x
2
– x – 2 – (3x
2
– 7) i.e.
y = 5 – x. The roots of the equation 3x
2
– 7 = 0 are the x
values at the points of intersection of the two graphs.
Now get graph papers and solve the two remaining equations.
Let your teacher check your work.
Note that to solve one graph using the other, the former graph
must rst be rearranged so that it equals to 0. The quadratic
expression of this arranged graph is then subtracted from the
quadratic expression of the drawn graph not vice versa. Usually,
the result of subtraction is a linear graph which is then drawn
using any three x values in the given range.
Exercise 15f
1. Using the scale of 2 cm to represent 2 units on the x axis
and 2cm to represent 1 unit on the y axis, draw the graph of
y = x
2
+ 5x – 6 for –6 x 1.
Use the graph to solve the equations
(a) x
2
+ 5x −6 = 0
(b) x
2
+ 5x − 6 = −10
(c) x
2
+ 2x − 7 = 0
2. Solve graphically the simultaneous equations y = x
2
and
y = x + 3 by drawing the two graphs on the same axes. Use
the scale of 2 cm to represent 2 units on both axes.
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3. On the same axes and scale, draw the graphs of
y = x
2
– 3x + 10. Use this graph to solve the equation
x
2
– 4x + 3 =0 for integral values of x from -1 to 5
4. On the same axes and scale draw the graph of
y = 2x
2
− 3x − 7 for the following values of x : −2 , −1 , 0, 1 ,
2 , 3, 4, 5. Use this graph to solve the following equation 2x
2
-5x − 6 = 0
5. By drawing the graphs of y = 2x
2
+ x – 3 and y = 2x – 1 on
the same axes, solve the equation 2x
2
+ x −3 = 2x – 1
6. Draw the graph of y = 4 + 3x – 2x
2
for values of x from −3 to
5. Use your graph to solve equation 5 + x – 2x
2
= 0
Activity 5:
Formulating quadratic equation from roots
Study the solution to the quadratic equation x
2
– x – 2 =0 below:
x
2
– x – 2 = 0
(x
2
– 2x) (+ x – 2) = 0
x(x – 2) + 1(x – 2) =0
(x + 1)(x – 2) = 0
x = –1 or x = 2
Now suppose you are given the roots (−1 and 2) of the quadratic
equation, discuss in groups how you can nd the quadratic
equation x
2
– x – 2 = 0. Report your ndings.
To nd the quadratic equation x
2
– x – 2 = 0, you will have to
work backwards from the roots.
Example 4:
Equation from roots
Formulate a quadratic equation whose roots are
and -
.
Solution:
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Let x =
or x = -
i.e. 3x = 2 or 5x = –1
i.e. 3x – 2 = 0 or 5x + 1 = 0
i.e. (3x – 2) ( 5x + 1) = 0
15x
2
– 7x – 2 = 0
Exercise 15g
Formulate a quadratic equation whose roots are:
a. 2 and 4
b. –1 and –5
c. –4 and
d.
- and -
e.
and -2
Activity 10:
Formulating the quadratic equation given a quadratic
graph which cuts the x-axis
Working in groups,
1. Draw the graphs of y = x
2
– x – 12 and y = 2x
2
– 2 x – 24 on
the same axes.
2. From your graphs what are the roots of each of the two
equations?
3. Now use the method of formulating the quadratic equation
given roots in activity 4.
4. Are you able to get both equations?
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You may have discovered that there are more than one
equation to the given roots. In fact, there are more than
two equations whose roots are 4 and -3. You should also
discovered that the method of working backwards works
only where the product of the roots equals the y-intercept of
the graph. Where the product of the roots is not the same as
y intercept of the graph we need another method.
Example 8
Find the equations of the graphs in the diagram below:
-15
Solution
Since –1 x 5 = –5, the equation of the graph which cut the y axis
at -5 is
(x + 1)(x – 5) = 0 i.e. x
2
– 4x – 5 = 0.
Now –15 is 3x – 5, hence the equation of the other graph is
3(x
2
– 4x – 5 = 0) i.e.
3x
2
– 12x – 15 = 0.
Example 9:
Finding equations of graphs
Find the equations of the graphs in the diagram below:
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- 5
2
-10
-15
(a)
(b)
Note that x = −5 and x = 2 are the x-intercepts.
Since −5 x 2 = −10, the equation of graph (a) is
y = (x + 5) (x − 2) = x
2
+ 3x − 10.
For graph (b), the y intercept is
or
or times the y intercept
of graph (a) so the equation of graph (b) is y =
(x
2
+3x +10)
i.e.
x
2
Example 10:
Finding equations of graphs
Find the equation of the parabola which has roots x = −2 and
x = -6 and cuts the y axis at y = 4
Solution
−2x −6 = 12
Now 4 =
times 12
Hence the equation of the parabola is y =
( x + 2)(x + 6)
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y = x
2
+ 8x + 12
= y x
2 +
x + 4
Exercise 15h:
Find the equation that denes each of the following graphs.
-3 6
-18
(a)
(c)
(b)
1. Find the equation of the parabola which has
a. Roots x = 3 and x = − 4 and its y-intercept is −12.
Find the equation of the parabola.
b. x-intercepts 3 and 4 and its y-intercept is − 6. Find
the equation of the parabola.
c. has x-intercepts 4 and −3 and its y-intercept is 24.
Find the equation of the parabola.
2. One of root of a quadratic equation is x = 0.5. The graph has
a turning point at (2, 3.5). Find the equation of the parabola.
3. Two quadratic graphs cut the x axis at x = -2 and at x = 5.
The y-intercept of the rst graph is 5 and the y-intercept
of the second graph is −10. Find the equation of the second
graph. (y = −1/2x
2
+ 1. 5x + 5
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Prepare your own questions on nding the equations of
parabolas which cut the x-axis.
Unit summary
In this chapter, you have learnt to draw and interpret graphs
of quadratic equations, how to solve graphically quadratic
equations and linear equations and how to formulate quadratic
equations given roots and given graphs which cut the x axis. You
also learnt to nd minimum and maximum value of a quadratic
function.
Unit review exercise
1. Draw the graph of y = x
2
– 3x + 2, taking the values of
x between o and 4. Use a scale of 2cm to represent to
represent 1 unit on the horizontal axis and 1cm to represent
1 unit on the vertical axis. From the graph, what is the
minimum value of the quadratic function?
2. By drawing the quadratic graph, nd the maximum value
of the quadratic function y = 1 – 2x – 3x
2
. Draw the graph in
the range of x values from -3 to 3 and using a scale of 2cm to
1 unit in the x axis and 1cm to 1 unit in the y axis.
3. Solve graphically the equation x
2
– 4x + 7 = x + 1. Use
2cm to represent 1 unit in the x axis and 2cm to represent 5
units in the y axis.
4. Draw, using the same scales and axes, the graphs of
y = x + 3 and y = x
2
– x + 1 for values of x from –3 to +4. Use
the graph to solve the following equations:
(a) x
2
–2x – 2 = 0
(b) x
2
– x – 2 = 0
5. By drawing the graphs of y = –x
2
+ 4 and y = x + 2 solve the
equation
−x
2
+ 4 = x + 2.
6. The equation y = – 0.035x
2
+ 1.4x + 1 where x and y are in
measured in cm is a model of the path taken by a bullet
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red from the ground level. What is the maximum height
reached by the bullet?
7. Members of a science club at a secondary school launch a
model rocket from ground level and the velocity of the rocket
is given by the formula h = −16t
2
+ 96t −128 where h is the
height in metres and t is the time in seconds. After how
many seconds will the rocket reach 128m above the ground?
8. A Blantyre City Council reghter aims a hose at a window
25m above the ground . The equation y = 5 + 2x – 0.05x
2
describes the path of the water.
How far from the building is the reghter?
Glossary
Line of symmetry: A line that divides the quadratic graph into
two congruent halves.
References:
1. Rheta N. Rubenstein et al ( 1995), Intergrated Mathematics
McDoughtal Little, New York
2. Larson, Kanold, Stiff (1998), Heath Algebra 2, Heath and
Company, New York
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Unit
16
INEQUALITIES
In your JCE mathematics, you
learnt how to show inequalities
on a number line. You also learnt
how to formulate and solve
inequalities.
In this unit, you will learn to
graphically present inequalities,
how to graphically illustrate
simultaneous linear inequalities,
and how to graphically illustrate
the solution to simultaneous
inequalities in two variables. You
will also learn how to present
inequalities in two variables
graphically.
Inequalities are used in modeling
real life situations such as
those that involve area, nding
dimensions of rectangles,
triangles and other geometrical
shapes. They are also used in
nding prots in a business.
Presenting inequalities
graphically
Activity 1:
Sketching linear inequalities in
one variable graphically
In pairs, discuss the following
questions:
1. What do you think a linear
inequality in one variable is?
2. Write at least ve examples
of linear inequalities in one
variable and let your teacher
check your answers.
A linear inequality in one variable
is an expression that contains one of
the four inequality symbols and has
its one variable raised to the power of
1. The examples of linear inequalities
in one variable are x > 5, x ≤ 2.5, x
> 0, y ≤ 5, y > 3, x ≤ 1.5. To sketch a
linear inequality in one variable you
use either a dashed line or a solid
line. A dashed line is used when
the inequality symbol is > or <. The
dashed line shows that the number
through which the line passes is not
included in the solutions. When the
symbol is ≤ or ≥ use the solid line to
show that the number through which
the line passes is included in the
solution set.
Example 1:
Present the graph of the following
inequalities on xy-plane.
(a) x > 5
(b) y ≤ 3
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Solutions
To present the above graphs, rst replace the inequality symbol
by “ = ” and draw the line of the resulting equation. Note that
the line will depend on the inequality symbol in the given
question. Then identify the side of this line containing the
solutions and shade the unwanted side of the line. The graphs
are shown below:
(a)
x = 5
y
x
0 5
(b)
y=3
3
Example 2:
Graph the solution of x –2 > 0, where x is a real number in the
xy -plane.
Solution
x – 2<0
∴
x< 2 + 0
∴
x < 2
y
x
0
2
x = 2
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Exercise 16a
Sketch the graphs to show the region represented by the
following inequalities in the xy-plane.
1. x < 4
2. y < –2
3. y ≤ –1
4. x + 2 > 5
5. 3 < 7 – x
6. x − 2 > 1
7. y ≤ 4
8. y ≤ 5
Illustrating simultaneous linear inequalities graphically
Simultaneous linear inequalities are inequalities composed of
two inequality statements in one sentence. They may be in one,
two or even in three variables. In this book you shall only look at
the rst two.
Activity 2:
Illustrating simultaneous linear inequalities in one
variable graphically
1. In your groups discuss the meaning of simultaneous linear
inequality in one variable.
2. Give examples of simultaneous linear inequality in one
variable.
Example 3:
Sketching graphs of inequalities
Sketch the graph to show the region represented by the
inequality −1 < x ≤ 3, in the xy-plane.
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Solution:
Note that this inequality is composed of two inequalities
−1 < x or x > −1 and x ≤ 3. Inequalities like these are called
simultaneous linear inequalities in one variable.
y
x
-1
0
3
x = 3
x = -1
Exercise 16b
Sketch the graph to show the region represented by the
following inequalities in the xy-plane:
1. 2< x ≤ 4
2. 4≤ x ≤ 5
3. 1< y < 4
4. 3≤ y< 3
5. 1 ≤ y ≤ 3
6. 2< x < 3.5
7. 1 < x < 2
8. 2.5 <y < 3.5
Activity 3:
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Illustrating linear inequalities in two variables graphically
1. Discuss in groups what linear inequalities in two variables
are. Give examples of these inequalities.
You are given four set of points in the xy – plane above
and the line whose equation is y = –x + 4 drawn from an
inequality y ≤ –x + 4. Note that this line divides the plane
into regions called half planes. By making substitutions into
the inequality, nd out which of the four sets of solutions
satisfy the inequality. Complete the table below:
Inequality Set of points Satises/doesn’t satisfy the
inequality(write true/not true)
y ≤ –x + 4
(-4,6)
True
y < –x + 4
(-4,-6)
y >
–x + 4
(4, 6)
y ≥
–x + 4
(8, 2)
Not true
2. Make a general statement about shading the unwanted side
basing on the results you found in 2.
3. Try other points and verify that the statement you have
made in 2 is correct.
4. Now replace the symbol ≤ in the inequality by one other
symbol and make substitutions as in 1, each time recording
which set of solutions satises the new inequality.
You should have seen that for the inequities y ≤ –x + 4 and y
< –x + 4,the set of points are true only for the points below the
line y = –x + 4. This means all the points below this line will
satisfy the inequalities. Hence the region above this line is an
unwanted region.
For the inequalities y > –x + 4 and y ≥ –x + 4, the set of points
are not true only for the points below the line y = –x + 4.
This means all the points below this line will not satisfy the
inequalities. Hence this region is an unwanted region. In other
words, for the inequities y > –x + 4 and y ≥ –x + 4, the wanted
region is above the line y = –x + 4.
You can now generalise the results as follows:
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As long as y (or any variable representing the vertical axis) is on
the left side of the inequality statement and is positive,
1. Shade above the line if the inequality symbol is ≤ or <.
2. Shade below the line if the inequality symbol is ≥ or >.
If y is on the right side of the inequality symbol, you may need to
rearrange the inequality for you to apply the results above.
Example 4:
Illustrate graphically the region represented by the
following inequalities on the xy plane:
(a) x + y < 4
(b) x + y ≤ 4
Solutions:
(a) • First know where the line x + y = 4 will cross the
x − axis by putting y = 0 in the equation i.e. x + 0 = 4,
∴
x = 4
• Also know where the line x + y = 4 will cross the
y − axis by Putting x = 0 in the equation i.e. 0 + y = 4,
∴
y = 4.
• You draw the graph of x + y = 4 as a dashed line
because the Inequality symbol is <.
• As y is on the left side of the inequality and is
positive, you shade above the drawn line (Rule 1
activity 3)
Graphically the line will look like this
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(-4,6)
(-4,-6)
y = -x + 4
(8,2)
-10
-8
-6
-4
-2
0
2
4
6
8
10
-10
-8
-6
-4
-2
2
4
6
8
10
(4,6)
In gure above, all points in the unshaded area satisfy the
inequality
x + y < 4.
(b) Similarly, the line to be used for the inequality is x + y ≤ 4.
The only Difference is that the line will not be dashed but a
solid line because the boundary is included in the
The graph will look like this.
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-10
-8
-6
-4
-2
0
2
4
6
8
10
-10 -8 -6 -4 -2 0 2 4 6 8 10
x + y < 4
Example 5:
Graph the inequality 2x + y > 6
Solution:
(a) Lets know where the line 2x + y = 6 crosses the x and y axes
by putting y and x = 0 respectively into the equation
2x + y = 6
2x + 0 = 6
2x = 6
x = 3
∴
I
t will cross the x axis at x = 3
Similarly putting x = 0
2
×
0 + y = 6
0 + y = 6
y = 6
∴
It will cross the y-axis at y = 6
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From result 2 of activity 3, you shade below the line 2x + y = 6.
The graph will look like this:
2x + y > 6
All points in the unshaded area satisfy the inequality 2x + y > 6.
Example 6:
Graph the inequality x – y ≤ 3
Solution:
For line x – y = 3
If x = 0, then y = −3
If y = 0, then x = 3
∴
T
he line will cross points (0, −3) and (3, 0)
Also note that y is on the left side of the inequality and is
negative. Rearranging, you shall have y ≥ 3 – x and so you shade
below the line.
Graphically, it will look like this:
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x − y ≤ 3
Exercise 16c
Graph the following inequalities in the xy-plane:
1. x + 3y < 6 2. 2x + y > 4
3. x + y > 5 4. 2x − y ≥ 4
5. x − y ≥ 0 6. −3x − 2y < −6
7. x − 2y < 0 8. −2x − 2y > 0
9. 3x − 5y ≤ 15 10. x + y ≥ 2
Illustrating simultaneous linear inequalities in two variables
graphically
Two or more inequalities may be graphed in the same coordinate
plane. The intersection of these graphs is the solution set for
the system of inequalities. In activity 3, you learnt how to graph
linear inequalities in two variables graphically. This is the same
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method used to illustrate simultaneous linear inequalities in
two variables.
Example 7:
Show graphically the solution set of the system y < x and
x + 2y ≤ 4.
Solution
Using ideas of activity 3 you come up with the following graph:
-1
0
1
2
3
4
5
-2 -1 0 1 2 3 4 5
y = x
x + 2y = 4
The solution set for y < x and x + 2y ≤ 4 is the unshaded area.
Exercise 16d
Graph each of the following simultaneous linear inequalities:
1. x + y ≥ 2 and x – y < 2
2. x + y ≤ 3 and x − y < 3
3. x ≥ y and x + y ≤ 6
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4. −2x − 3y > −6 and 3x – 2y > 6
5. y ≥ 2 x and y ≥ −x
6. 2x − 5y ≤ 10 and x + y ≤ 5
7. y − x≤ 2 and x + y < 1
Writing down inequalities that describe a given region
To write down an inequality describing a given region, you need
to consider a number of things;
• whether the line dividing the plane into half planes is
dashed or solid.
• which side of the line is the wanted side or contains the
solutions of the inequalities.
• the equation of the line dividing the plane into half planes.
Example 8:
Inequalities for a given region
Write down the inequalities that describe the unshaded regions
below;
y
y
4
2
x
x
-3
-2
0
4
0
Solutions
(a) The line passes through x = –3 so its equation is x = –3. The
line is dashed so the inequality symbol must be < or >. The
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wanted side is to the left of the line where numbers are less
than -3. Hence the inequality is x < –3.
(b) You must note that in this example there are three lines
bordering the unshaded region. The gradient of the line
passing through (–2, 0) and (0, 2) is
= 1
. Therefore its
equation is y – 0 = 1(x– –2) i.e. y = x + 2. The line is solid
so the inequality symbol must be ≤ or ≥. The line is shaded
above and in the equation, y is on the left and is positive.
Therefore the inequality is y ≤ x +2 (activity 3).
By a similar method, the equation of the line passing
through (4, 0) and (0, 4) is y = –x + 4. The line is dashed so
the inequality symbol must be < or >. The line is shaded
above and y is on the left of the equation and is positive
therefore the inequality is y ≤ –x + 4.The equation of the
third line is y = 0 and is shaded below. Therefore the
inequality is y ≥ 0.
Exercise 16e
Write down the inequalities that describe the unshaded region
in each of the following:
y
(1) (2) y
x x
0 2 0
-3
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y
y
(3)(4)
3
x
x
-1 0 3 -2 0 3
-4
(5)
5
3
-4 0 4
-
-4
Unit summary
In this unit, you have learnt to present inequalities in one and
two variables graphically. You have also learnt to illustrate
graphically the solutions to simultaneous linear inequalities
in one and two variables and writing down inequalities that
describe a given region.
Glossary:
Wanted region: The region that contains solutions to an
inequality or a set of inequalities.
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Linear inequalities: Inequalities whose equations produce a
straight line graph.
Unit review exercise
1. Draw diagrams to represent the following inequalities:
1. x ≥ 2
2. y ≥ –3
3. x < 5
4. y < 2
5. –2 < x < 5
6. 2 < x + y ≤ 5
2. Write down the inequalities that describe the unshaded
regions in each of the following:
(a)
y
3
-3 0 4 x
-2
(b)
y
3
0 x
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(c)
-3
y
2
0 2 x
(d)
y
- 2 0 x
-3
(e)
4
2
0 3 5 x
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Unit
17
STATISTICS
In your JCE Mathematics you
learnt about some statistics.
You learnt about collecting and
classifying and presenting data
in form of graphs.,In this unit,
you will learn about organisation
of data, presenting data in form
of charts and tables, calculating
measures of central tendency and
spread and interpreting data.
Organisation of data
When you rst look at some data, all
you can see is a jumble. Such data
is called raw data or unprocessed
data. Raw data is hard to analyse.
Therefore data must be organized
in such a way that it can easily
be understood. In the activities
that follow, you will learn how to
organize data so that it can easily be
understood.
Activity 1:
Classifying grouped data
There are two types of data:
Qualitative and Quantitative data.
Qualitative data are non-numerical
data e.g. the texture and colour of a
fabric.
Quantitative data are numerical
data e.g. the numbers of people in a
room or the height of a person.
Quantitative data is further classied
into two groups: Discrete and
Continuous data.
Discrete data are countable for
example: The number of rooms
in a house or a person’s shoe size.
Continuous data are measured
data, for example: length, weight,
temperature and time are all
measured on a continuous scale.
Now in pairs, discuss whether
the following data are discrete or
continuous: Explain your answers.
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1. Your weight from birth to age 14.
2. The time you get up each morning for one month.
3. The number of mangoes you sell at the market each week.
4. The number of peas in a pod.
5. The age of pupils at your school.
Example 1:
Classifying data
Classify the following groups of data as discrete or continuous:
a. a group of 16-20 years of age.
b. a group of students who attended lessons for 15-30 periods
in a week.
Solutions
a. It is continuous data because between any two weights you
measure any value is possible.
b. It is discrete data because number of periods is distinct. You
cannot have one and half periods or two and three-quarter
periods e.t.c.
Exercise 17a
Classify the following grouped data as discrete or continuous.
The data properties or what the data stand for is shown in the
brackets against each group:
1. 50 – 80 (marks in %)
2. 20 – 30 (temperature in degrees Celsius)
3. 40 – 60 (speed in km/h)
4. 61 – 70 (heights in cm)
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5. 100 – 150 (weight in kg)
6. 10 – 15 (Number of rooms in houses)
7. 4 – 6 (Shoe sizes)
8. 2 – 10 (number of people in cars)
9. 0 – 9 (number of absentees)
10. 19 – 20 (age group)
11. 0 – 99 (monthly salary in Kwacha)
Class intervals
In many cases you may have a set of many items e.g. numbers.
It is hard to make an analysis of the numbers or it is hard to
make any interpretation from the set. To get a clearer picture of
the data you group the data within class intervals. Each class
interval has the beginning and end. The beginning is called the
lower limit and the end is called the upper limit. The number
of items from the lower limit to the upper limit is called class
width.
The following things need to be considered when grouping data
within class intervals:
1. The starting point of the rst class interval must be the
smallest number in the given set or a number just below
that.
2. Each class interval must be of the same class width.
3. The class intervals must not overlap.
4. The last class interval must contain the highest number in
the given data.
Activity 2:
Forming class intervals
You are given a set of different numbers representing marks
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scored by students in a mathematics test marked out of 100.
The lowest mark is 1 and the highest mark is 100. In pairs form
class intervals given that
(a) The rst class interval is 1 – 10.
(b) The rst class interval is 1 – 20.
Compare your work with that from other pairs.
Suppose the rst class interval was 1 – 5, why would the
intervals 1 – 5, 5 – 9, 9 – 13 e.t.c. be wrong?
Example 2:
Class intervals
The table below gives masses, in kg, of 30 students:
43 45 50 47 51 58 52 47 42 54
61 50 45 55 57 41 46 49 51 50
59 44 53 57 49 40 48 52 51 48
Table 1
Form class intervals with 40 – 45 as the rst class interval.
Solution
First note that this is continuous data and as such the class
intervals must cover all possible masses of data. Hence you can
have the following intervals: 40 – 45, 45 – 50, 50 – 55, 55 – 60
and 60 – 65.
Example 3:
Class intervals
The number of people passing through a check point was
recorded after every 30 minutes for 24hours. The smallest
number recorded was 3 and the highest number recorded was
40. Form class intervals for the data with 1 – 5 as the rst
interval.
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Solution
Unlike in example 1, this is discrete data and so there are no
half values between the class intervals. Hence the intervals are
1 – 5,
6 – 10, 11 – 15, 16 – 20, 21 – 25, 26 – 30, 31 – 35, and 36 – 40.
Exercise 17b
In questions 1 – 5, the rst class interval and the class of data
are given. Form the next 5 class intervals for each question.
1. 1 – 9 ;(discrete data)
2. 1 – 15; (discrete data)
3. 40 – 45; (continuous data)
4. 20 – 30; (continuous data)
5. 40 – 59; (continuous data)
In questions 6 – 10 the lower limit of the rst class interval,
the class width and the class of data are given. Form 5 class
intervals for each question.
6. Lower limit = 1, class width = 5, discrete data.
7. Lower limit = 10, class width =10, discrete data.
8. Lower limit = 0, class width = 20, discrete data.
9. Lower limit = 50, class width = 5, continuous data.
10. Lower limit = 8, class width = 6, discrete data
11. 50 cars were tested to see how far they had travelled on 10
Litres of certain petrol and the distances travelled in km were
recorded as shown in table 2 below:
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Table 2
100 110 130 120 140 121 127 142 126 143
130 134 107 131 145 111 146 112 147 144
131 130 132 129 108 132 131 125 149 140
128 106 133 128 128 148 123 133 141 122
135 139 134 122 143 101 128 138 121 149
The rst two of the ve class intervals for the data are 100 – 109
and 110 – 119. Write down the other intervals.
Activity 3:
Determining class boundaries
A class boundary separates one class from the other i.e. the
ending point of one group is the starting point of the other. In
groups, discuss how you can determine the class boundaries for
the following class intervals: The data is continuous.
1 – 9, 10 – 19, 20 – 29, ---
Present your work to class.
Each class boundary lies halfway between the upper limit stated
in one class interval and the lower limit stated in the next. This
means the class boundaries are the averages of the two limits in
the intervals given. The number of values from the lower limit to
the upper of each class gives the class width.
Example 4:
Class boundaries
The following are intervals of temperature measured in degrees :
1 – 5, 6 – 10, 11 – 15 … Determine the class boundaries for the
groups.
Solution
Get the averages of the upper and lower limits for adjacent class
intervals as follows:
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, , = 0.5, 5.5, 10.5, ---
So the class boundaries are 0.5, 5.5, 10.5 and so on.
Example 5:
Class intervals
The following are class intervals of ages of a group of students.
0 – 9, 10 – 19, 20 – 29.
Determine the class intervals for the groups.
Solution
Ages are usually given in completed years. For example you say
someone is 9 years old from his or her 9
th
birthday until just
under their 10
th
birthday. This means someone who is 9 years
364days i.e. 10 years belongs to a group
0 – 9. So for ages, 0 – 9 means from 0 to just less than 10 years.
So the class boundaries for the groups are 0, 10, 20, 30.
Exercise 17c
1. The intervals below are weights (in grams) of a number of
tomatoes for grading purposes:
60 – 65, 65 – 70, 70 – 75, 75 – 80, 80 – 85.
Determine the class boundaries for the intervals.
2. The following are the intervals of lengths ( in mm) of 30
leaves measured by students in a statistics lesson:
30 – 40, 40 – 50, …, 90 – 100.
What are the class boundaries for the groups?
3. The ages of people attending a lesson on environment are
grouped as follows: 0 – 9, 10 – 19, 20 – 29, 30 – 39, 40 – 59,
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60 – 69 , 70 – 79.
Determine the class boundaries of the intervals.
4. The temperature, measured to the nearest degree Celsius,
recorded at a certain place over a period of time is grouped
as follows:
0 – 14, 15 – 29, 30 – 34, 35 – 49, 50 – 64, 65 – 79.
What are the class boundaries for the data?
5. Several books were collected to see how many pages each
book had. The pages of the books were put in the following
categories:
1 – 5, 6 – 10, 11 – 15, 16 – 20, 21 – 25 , 26 – 30.
Determine the class boundaries of the groups.
Activity 4:
Finding mid-points of class intervals
In pairs, discuss what “mid-point of a class interval” is.
Suppose 40 – 59 is a class interval, what will be the mid-point of
this Interval? Explain how you got it.
The mid-point of a class interval is the value midway between
the lower limit and the upper limit of each class interval. It
may be calculated by nding the arithmetic mean of the two
numbers.
Example 6
Find the mid-point of the interval 86 – 100.
Solution
Mid-point =
= 93
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Exercise 17d
Find the mid-points of the following class intervals:
1. 20 – 24
2. 72.8 – 72.9
3. 12 – 30
4. 1 – 10
5. 80 – 100
6. 30 – 39
7. 140 – 144
8. 0 – 99
9. 100 – 119
10. 40 – 50
Presenting data in forms of charts and tables
Diagrams and tables are often used to display data. They are
simpler to understand than just crumbled data. Diagrams
look more attractive and interesting and help you to spot any
patterns and compare things easily. In the activities that follow,
you will learn to present data in form of frequency tables,
histogram, pie charts and frequency polygons.
Activity 5:
Presenting data in the form of a frequency table
Frequency is how many times a value in the given data occurs.
Frequency of things can be shown in a table called frequency
table.
In your JCE you learnt how to tally the given data and draw
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frequencies from them. In pairs, discuss how you can come up
with a frequency table for the data below. Draw the table. The
data shows marks obtained by 25 students in a test.
14 19 9 17 15 20 17 10 15
12 17 11
17 15 16 17 19 17 12 8 17
10 12 15
18 15 10 7 10
From your table
(a) What is the total frequency?
(b) What are the frequencies of the following scores? 12,17 and
20 Present your ndings to class.
Exercise 17e
1. Consider the data below:
1 1 2 0 0 3 1 1 2
2 2 1 1 0 4 1 0 2
3 1 2 1 3 0 1 2 1
4 0 1 2 2 0 1 2 2
0 1 2 2 3 3 0 1 4
(a) Draw a frequency table for the distribution.
(b) What are the frequencies of the following: 0 , 4, 1?
Construct a frequency table for the distribution below:
1 1 3 3 3 0 2 0 1 2
0 3 0 4 2 3 0 3 1 3
3 4 0 0 6 3 5 0 1 0
3 1 0 7 4 3 3 0 2 3
1 3 3 2 5 0 3 0 6 3
1 0 3 5 0 1 4 2 3 4
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Which value ha the highest frequency?
3. Consider the scores below:
26 12 24 42 16 18 30 24 36 8
34 24 20 16 24 16 18 26 10 32
24 12 26 12 12 20 24 18 14 22
24 14 34 24 20 12 20 24 20 26
32 12 16 42 26 22 24 20 16 18
18 30 24 36 41 35 30 41 10 36
(a) In groups, form class intervals for the data using
intervals 5 – 9, 10 – 14, ---
(b) Draw the frequency table for the data.
(c) Which group has the highest frequency?
4. The following marks were obtained by 80 candidates in a
Mathematics test which was marked out of 65.
54 52 31 47 24 36 27 15 44 26
8 20 46 32 27 31 33 57 39 32
43 32 23 33 31 21 38 28 40 19
52 37 38 39 9 30 47 29 8 13
33 35 48 18 36 39 23 58 34 35
16 21 32 38 34 13 27 32 37 23
37 49 25 38 24 27 48 36 45 18
41 34 43 12 47 24 61 29 37 33
Using intervals 0 – 9, 10 – 19, ---, draw the frequency table
for the data.
Activity 6:
Presenting data in the form of the histogram
In groups,
1. Find the class boundaries of the following class intervals:
1 – 5, 6 – 10, 11 – 15, 16 – 20.
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2. Given that the frequencies of the above intervals are 2,3,5,7
draw a bar graph on a chart paper with the class boundaries
on the horizontal axis and frequencies on the vertical axis.
3. Find the area and the width of each bar.
4. Now for each bar, divide the area by the width. What do you
notice?
5. Report your ndings to class.
A histogram often looks like a bar chart that you dealt
with in your JCE. However the essential characteristic of
a histogram is that it represents frequency by area and
that there is a numerical relationship between quantities
whose frequencies it represents as you saw in instruction 5
in the above activity. In addition there are no gaps between
adjacent bars in the histogram.
Example 7:
Drawing histogram
The frequency distribution below shows weekly earnings of 50
people.
Weekly earnings
(Kwacha)
0– 99 100 – 199 200– 299 300– 399 400– 500
Number of people 5 16 19 6 4
Draw a histogram for the frequency distribution.
Solution:
As the rst interval starts from 0,using the method of nding
class boundaries shown in example 3 on page 359 would lead
into the rst class having a class width different from the others
so use the lower limits of the intervals as boundaries. The
histogram is shown below:
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0 100 200 300 400 500
20
15
10
Frequency
20
15
Frequency
10
5
Weekly earnings ( kwacha)
Example 8:
Drawing a histogram
The data below shows a frequency distribution of temperature
in degrees Celsius recorded at a weather station over a period of
time.
Class interval (t minutes) Frequency
1 – 5 5
6 – 10 13
11 – 15 14
16 – 20 12
21 – 25 6
Draw a histogram for the distribution.
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Solution
The class boundaries are 0.5, 5.5, 10.5, 15.5 and 25.5
20
15
10
Frequency
Frequency
0
0.5
5.5 10.5
Temperature (ºC)
15.5 20.5
25.5
5
10
15
20
Activity 7:
Presenting data in the form of a frequency polygon
If you draw straight lines through the mid-points at the top
of each bar of a histogram, the resulting graph is a frequency
polygon. Basing on this statement, in groups discuss how you
would construct a frequency polygon using the data provided in
example 8. Draw the polygon.
Example 9
Use the frequency table below to draw a frequency polygon.
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Class interval (Marks) 1 – 5 6 – 10 11 – 15 16 – 20 21 – 25
Frequency 3 5 7 6 4
Solution
The class boundaries are 0.5, 5.5, 10.5, 15.5, 20.5 and 25.5 .The
mid points of the class intervals are 3, 8, 13, 18 and 23. You then
draw a histogram for the data and then a frequency polygon as
below:
Frequency
5
0
0.5
5.5 8 10.5 13
Marks
15.5 18 20.5
23
25.5
3
10
If you only use the midpoints, you get the following frequency
polygon. Note that extra intervals have been added at the
beginning and at the end to remove “hangings” of the polygon.
10
5
Frequency
-2 3 8 13 18 23 28
10
5
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Exercise 17f
1. Draw a histogram hence a frequency polygon for the table
below:
Height, x cm Frequency
130 ≤ x < 135
135 ≤ x < 140
140 ≤ x < 145
145 ≤ x < 150
150 ≤ x < 155
8
14
17
12
4
Total: 55
2. The temperatures recorded one summer were as follows
Temperature
0
F 68 69 70 71 72 73 74 75 76 77 78
Number of days 3 5 12 14 13 8 5 9 6 3 2
Draw a histogram hence a frequency polygon.
3. The lengths of pupils’ forearms were recorded as follows:
Length of forearm (cm) 24 25 26 27 28 29 30 31 32 33
Number of pupils 8 9 21 24 20 27 16 22 12 9
Draw a histogram hence a frequency polygon.
4. The year of manufacturing 100 cars in a car factory is:
Year 1994 1995 1996 1997 1998 1999 2000
Number of cars 42 48 57 79 74 85 52
Draw a histogram for the data.
5. The table gives the time taken for students to travel to
school.
Time (minutes 0 ≤ t < 10 10 ≤ t < 20 20 ≤ t < 30 30 ≤ t < 40
Frequency 12 36 8 5
Draw a histogram for the data.
6. The actual time between departure of minibuses at rush
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hour was recorded to the nearest minute.
Time (minutes) 7 8 9 10 11 12
Number of minibuses 13 18 35 12 19 3
Draw a histogram for the data.
Activity 8:
Presenting data in a form of the pie chart
In your groups do a research on the following:
1. Denition of a pie chart.
2. The steps in drawing a pie chart.
Report your ndings to class.
A pie chart is a circular diagram used to display data. The whole
pie stands for the whole amount of data being dealt with and
each slice stands for a named part of the data. To draw a pie
chart you follow the following steps:
1. Find the total frequency of the given data.
2. Divide 360
0
by the total frequency.
3. Multiply each frequency by the result in 3. This gives the
size of each slice.
4. Measure the angles in 3 above anticlockwise at the centre of
the circle.
Example 10:
Drawing pie charts
Draw a pie chart to show the following daily life of Takondwa:
Activity Number of hours
Lessons 5
Meals 1
Homework 3
TV 2
Travel 1
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375
Sleep 8
Other 4
Solution
Total frequency = 5 + 1 + 3 + 2 + 1 + 8 + 4 = 24
Angle for 1 of each activity = 360
0
24
0
= 15
0
Angles for each activity are 5x15
0
, 1 x 15
0
, 3x15
0
, 2x15
0
, 1
x15
0
,8x15
0
and 4x15
0
i.e.75
0
, 15
0
, 45
0
, 30
0
, 15
0
, 120
0
and 60
0
The pie chart is shown below:
Meals
Homework
15
0
TV
45
0
Lessons
Travel 15
0
30
0
75
0
120
0
60
0
Sleep
Others
Exercise 17g
1. The frequency distribution below shows how a
manufacturing rm used its income in 2004:
Area of spending Wages Building Raw materials Prot
Amount spent (%) 5 15 75 5
Draw a pie chart to show the distribution.
2. A school allocates its weekly timetable to pupils in Form 3
as follows:
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375
Subject English Mathematics Physical science
Number of periods 12 15 18
Represent the above information in a pie chart.
3. The total surface area of the earth is approximately 510
million km
2
. The area is composed of water and land as
shown in the pie chart below:
29%
Water
What angle of the pie chart represent water?
4. In a village 324 litres of water are used each day. The pie
chart below shows how the water is used.
Cooking Washing
54
0
Given to Washing
animals clothes
How much water is used in washing clothes?
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Measures of central tendency
In your JCE you learnt about three measures of central
tendency, mean , mode and median. You learnt how to calculate
them from an ungrouped data. You will now learn how to
calculate the measures from grouped data.
Activity 9:
Calculating median of grouped data
In groups, discuss how you can nd the middle value (median)
from the frequency distribution below:
Age (years) 0-4 5-9 10-14 15-19 20- 24 25-29
Number of people 2 3 6 9 8 5
You cannot nd median directly from a frequency distribution
table. The table helps you to nd the class that contains the
median (median class).
You add the frequencies cumulatively to nd the position of the
middle value(s). If the sum (n) of frequencies is even, the median
is
th
and ( + 1)
th
values while if the sum of frequencies is odd
the median is
th
value.
Example 11
Find the median of the frequency distribution below.
Class interval 0-9 10-19 20-29 30-39 40-49 50-59
Frequency 1 2 11 9 14 3
Solution
Sum of frequencies = 1 + 2 + 11 + 9 + 14 + 3 = 40
Hence the middle values are 20
th
and 21
st
values. Add the
frequencies cumulatively as follows:
1 + 2 = 3 values
3 + 11= 14 values
14 + 9 = 23 values
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377
Hence the 20
th
and 21
st
values are in the class in the class
interval 30-39. Hence 30 - 39 is the median class.
But if we need to estimate a single Median value we can use
this formula:
Estimated Median = L +
(n/2) – cf
b
× w
f
m
where:
• L is the lower class boundary of the group containing the
median
• n is the total number of data
• c f
b
is the cumulative frequency of the groups before the
median group
• f
m
is the frequency of the median group
• w is the group width
So the estimated median of the above data
=30 +
x 10
= 30 + 6.7
= 36.7
Exercise 17h
1. The frequency distribution below shows the heights in cm of
50 students.
Height (cm) Frequency
120 -129 8
130-139 3
140-149 9
150-159 22
160-169 7
170-179 1
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a. What is the median class of the data?
b. Estimate the median of the data.
2. The following are the weights 30 pupils in kg.
45 62 35 54 48 35
48 59 52 40 54 46
59 51 32 37 49 42
53 38 37 35 53 46
48 44 33 52 54 44
Construct a frequency distribution using intervals 30 – 39, 40 –
49,--- and nd the median class and the median of the data.
3. The frequency distribution below shows number of students
who scored marks within 10-mark class intervals in a test.
Marks Frequency
1-10 2
11-20 7
21-30 9
31-40 11
41-50 13
a. Find the median class of the distribution.
b. Calculate the median of the data.
4. Find the median of the frequency distribution which shows
diameters in mm of 24tins.
Diameter(mm) 60-69 70-79 80-89 90-99 100-109
Frequency 1 3 7 8 5
5. Find the median class and median of the frequency
distribution below.
Class interval 31 - 50 51-70 71-90 91-110
Frequency 1 11 4 3
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Activity 10 :
Calculating mean of grouped data
The table below is given in example 1 of this topic.
43 45 50 47 51 58 52 47 42
54 61 50 45 55 57 41 46 49
51 50 59 44 53 57 49 40 48
52 51 48 1
In groups,
1. Calculate the mean of the above data using the method you
learnt in form1 to the nearest whole number.
2. Group the data in the following intervals:
40 – 44, 45 – 49,---.
Construct a frequency table using intervals for the data
using the intervals.
3. Find the midpoint of each class interval.
4. Multiply each midpoint by the frequency of that group.
5. Sum up the frequencies and the products.
6. Divide the sum of all the products by the total frequency
correcting their answer to the nearest whole number.
7. Comment on your ndings.
The method of nding mean by grouping data only gives an
approximation but is a quicker and less tiring way of nding the
mean where the data is so large. The method can be summarised
into the following steps for nding mean of grouped data:
1. Find the frequency and midpoint of each class interval.
2. Multiply each midpoint by the frequency of that group.
3. Sum up the frequencies and the products.
4. Divide the sum of all the products by the total frequency.
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Example 12
Calculate the mean of the frequency distribution below.
Number of cars 10-14 15-19 20-24 25-29 30-34 35-39 40-44 45-49
Number of days 1 1 2 8 19 14 4 1
Solution
Class interval Midpoint of class interval Frequency Midpoint x Frequency
10-14 12 1 12
15-19 17 1 17
20-24 22 2 44
25-29 27 8 216
30-34 32 19 608
35-39 37 14 518
40-44 42 4 168
45-49 47 1 47
Sum=50 Sum= 1630
Mean = 32.6 cars/day.
Sometimes you can use what is known as a working mean. A
working mean is the central value of the class with the highest
frequency in this case 30 -34. The central value is 32. You then
proceed as follows:
Class interval Class central
Value(x)
Frequency (f) Central value-Working
mean. This is a called
deviation.
Frequency x
Deviations
10-14 12 1 −20 −20
15-19 17 1 −15 −15
20-24 22 2 −10 −20
25-29 27 8 −5 −40
30-34 32 19 0 0
35-39 37 14 5 +70
40-44 42 4 10 +40
45-49 47 1 15 +15
Total deviations 30
Mean deviation = 30 ÷ 50 = 0.6
Then add this mean deviation to the working mean.
Mean of the data = 32 + 0.6 = 32.6 cars/day.
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Exercise 17i
Find the mean of each of the following frequency distributions:
1.
Class interval 1- 5 6-10 11-15 16-20 21-25
Frequency 3 5 10 6 4
2.
Class interval 21-30 31-40 41-50 51-60 61-70
Frequency 2 5 7 9 11
3.
Class interval 100-109 110-119 120-129 130-139 140-149
Frequency 5 15 25 35 20
Find the working mean and hence the mean of the following
frequency distributions:
4.
Class interval 0-4 5-9 10-14 15-19 20-24
Frequency 5 20 5 10 10
5.
Class interval 0-49 50-99 100-149 150-199 200-249
Frequency 6 8 11 9 4
Activity 11:
Calculating mode of grouped data
You cannot give the exact value of mode from grouped data.
However, you can make a reasonable estimate of it by using the
formula below.
Estimated Mode = L +
f
m
− f
m-1
× w
(f
m
− f
m-1
) + (f
m
− f
m+1
)
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383
Where:
• L is the lower class boundary of the modal group( the group
with the highest frequency)
• f
m-1
is the frequency of the group before the modal group
• f
m
is the frequency of the modal group
• f
m+1
is the frequency of the group after the modal group
• w is the group width
Go back to activity 10 above and nd the modal class and the
mode of the data.
1. Using the method of mode of ungrouped data nd the mode
of the data in activity 9 and compare it to the mode you
have just found.
2. Comment on your ndings.
Example 13
The following frequency table was drawn up for the marks in a
mathematics test.
Class Frequency
50-53 7
54-57 8
58-61 9
62-65 6
66-69 4
70-73 1
a. What is the modal class?
b. Estimate the mode of the data
Solution
a. Modal class = 58-61
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383
b. Mode = 58 +
( )
x 3
= 58 + 0.2
= 58.2
Exercise 17j
For each of the following frequency distributions nd the modal
class and hence estimate the mode.
1.
Class interval frequency
0 – 9 3
10 – 19 5
20 – 29 11
30 – 39 9
40 – 49 8
2.
Class interval Height (cm) frequency
40 – 44 7
45 – 49 9
50 – 54 10
55 – 59 6
60 - 64 2
3.
Mass of nails (g) 0-4 5-9 10-14 15-19 20-14
Number of nails 15 12 10 5 3
4.
Length of life (hours) 201-300 301-400 401-500 501-600 601-700
Number of bulbs 10 16 32 54 88
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Activity 12:
Calculating the range of grouped data
In groups, discuss how you can get the range from the frequency
distribution in example 13.
In a grouped frequency distribution, you can also make a
reasonable estimate of the range. It is found by subtracting the
lower limit of the rst class interval from the upper limit of the
last class interval.
Example 14
Estimate the range of the data in example 13.
Solution
Range = 73 – 20 = 23.
Exercise 17k
Estimate the range of the data in question 1 to question 4 of
exercise 17i.
Unit summary
• In this unit you have learnt about organisation of data,
presenting data in form of charts and tables, calculating
measures of central tendency and spread.
• In this unit you have also seen that you cannot give exact
values of mean, mode and median. You can only give
estimates of them.
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385
Unit review exercise
1. The rst class interval of discrete data is 0- 4. Form the next
5 class intervals.
2. The lower limit of the rst class interval of continuous data
is10, the class width is 8. Form the rst three class intervals
for the data.
3. Two rst class intervals of data are 15-19 and 20-24. Find
(a) the class boundary and(b) the midpoint of the class
intervals.
4. The weights ( in kg correct to the nearest tenth) of a group
of thirty dogs were recorded as follows:
15.5 14.8 15.8 14.3 14.6 15.0
16.2 13.9 15.2 15.1 16.0 15.2
14.4 15.4 15.7 16.2 14.9 14.7
15.5 13.7 15.5 14.3 14.7 15.1
Represent the data in the form of a frequency distribution
using intervals 13.5 -13.9,14.0-14.4,---
5. 45 children working in groups, were asked to time each
other’s estimates of the length of a minute. Their estimates
correct to the nearest second , are given below.
53 47 77 63 59 54 62 65 71
77 42 68 67 51 72 57 73 48
61 46 51 50 63 68 54 50 65
53 78 69 44 56 77 58 55 79
66 58 67 52 48 70 49 71 73
Using intervals 40 – 44, 45 – 49,---, draw a histogram for the
data.
6.
Number of games Frequency
1 - 5 2
6 - 10 7
11 - 15 8
16 - 20 3
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387
Using the frequency distribution above nd
a. the mode
b. the median
c. the modal class
d. the median class
e. the range.
f. the mean.
Glossary
Qualitative data: Data of properties that are numbers.
Quantitative data: Data of properties that are not numbers.
Discrete data: Countable data.
Continuous data: Measured data
Lower limit: The beginning of one class interval.
Upper limit: End of class interval.
Class width: Number of values from the lower to the upper
limit.
Class boundary: A point where one class ends and the other
begins.